Free Math Worksheets

- What does the first derivative represent geometrically? - What is the slope at the highest point? - Substitute the critical value into the original function. - Interpret a zero derivative in terms of the tangent line.

1. Differentiate: \(h'(x) = -0.1x + 0.8\). 2. Solve \(h'(x) = 0\): \(-0.1x + 0.8 = 0\), so \(x = 8\,\text{m}\). 3. Since \(h''(x) = -0.1 < 0\), this critical point gives a maximum. 4. Evaluate \(h(8) = -0.05(8^2) + 0.8(8) + 1.5 = 4.7\,\text{m}\). 5. The equation \(h'(x) = 0\) means the tangent line is horizontal. At this point, the path changes from increasing to decreasing.

a) \(8\,\text{m}\) b) \(4.7\,\text{m}\) c) The tangent line is horizontal, and the path changes from rising to falling.

- Relate the bottom width and the two side-wall heights to the total strip width. - Write the rectangular cross-sectional area in terms of \(h\) alone. - Maximize the resulting quadratic function. - Check that the answer is physically possible.

1. Let \(b\) be the bottom width. The strip-width constraint is \(2h+b=24\), so \(b=24-2h\). 2. The cross-sectional area is \(A(h)=h(24-2h)=24h-2h^2\), for \(0\le h\le 12\). 3. Differentiate: \(A'(h)=24-4h\). The critical point is \(h=6\,\text{cm}\). Since \(A''(h)=-4<0\), the area function is concave down, so this point gives the absolute maximum.

The side walls should be \(6\,\text{cm}\) high.

- What condition does the derivative satisfy at an interior maximum? - Which critical time is meaningful in the context of the problem? - After finding the time of the maximum, remember to evaluate the original height function.

1. Differentiate the height function: \(h'(t) = -0.3t^2 + 0.4t + 1.5\). 2. Set the derivative equal to zero: \(-0.3t^2 + 0.4t + 1.5 = 0\). The solutions are \(t = 3\) and \(t = -\frac{5}{3}\). Since \(t \ge 0\), only \(t = 3\) is relevant. 3. Use the second derivative: \(h''(t) = -0.6t + 0.4\), so \(h''(3) = -1.4 < 0\). Therefore, the height has a maximum at \(t = 3\). 4. Evaluate the height: \(h(3) = -0.1(3^3) + 0.2(3^2) + 1.5(3) = 3.6\). The maximum height is \(3.6\,\text{km}\).

\(3.6\,\text{km}\)

- What is the distance from a point \((x, y)\) to the x-axis? - Since the graph stays above the x-axis, which values of \(f\) give the smallest distance? - Find and classify the critical points of \(f\). - Give complete ordered pairs in your answer.

1. Because \(f(x) > 0\), the distance from a point \((x, f(x))\) to the x-axis is \(f(x)\). Minimize \(f\). 2. Differentiate: \(f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1)\). The critical numbers are \(x = -1\), \(x = 0\), and \(x = 1\). 3. Use \(f''(x) = 12x^2 - 4\). The points at \(x = -1\) and \(x = 1\) are local minima, while the point at \(x = 0\) is a local maximum. 4. Since \(f(x) \to \infty\) as \(x \to \pm\infty\), the local minima are absolute minima. Evaluate \(f(-1) = 4\) and \(f(1) = 4\). 5. Therefore, the closest points are \((-1, 4)\) and \((1, 4)\).

\((-1, 4)\) and \((1, 4)\)

- Marginal cost is the first derivative of total cost. - Which derivative locates an extremum of marginal cost? - An inflection point marks a change in concavity. - Compare how the slope behaves on each side of the inflection value.

1. The marginal-cost function is \(K'(x)=x^2-12x+45\). Its derivative is \(K''(x)=2x-12\). 2. Set \(K''(x)=0\): \(2x-12=0\), so \(x=6\). Since \(K'''(x)=2>0\), the marginal cost has a minimum at \(x=6\). 3. Evaluate the total cost: \(K(6)=72-216+270+120=246\). The inflection point is \((6, 246)\). 4. Before \(x=6\), the marginal cost is decreasing, so the total-cost graph is concave down. After \(x=6\), marginal cost is increasing, so the total-cost graph is concave up. Thus, \(x=6\) is the transition from decreasing marginal cost to increasing marginal cost.

a) \(x=6\) units b) \((6, 246)\), representing a total cost of \(\$246\). At this production level, marginal cost changes from decreasing to increasing.

- How do you form profit from revenue and cost? - What does a negative profit value mean? - Find the critical numbers of the profit function. - Verify that the candidate gives a maximum. - Compare with the endpoints of the allowed production interval.

1. Form the profit function: \(P(x) = 25x - (x^3 - 6x^2 + 10x + 50) = -x^3 + 6x^2 + 15x - 50\). 2. Evaluate \(P(2) = -8 + 24 + 30 - 50 = -4\). Because the units are thousands of dollars, this is a loss of \(\$4{,}000\). 3. Differentiate: \(P'(x) = -3x^2 + 12x + 15\). Solving \(P'(x) = 0\) gives \(x = -1\) and \(x = 5\). Only \(x = 5\) lies in the production interval. 4. Use \(P''(x) = -6x + 12\). Since \(P''(5) = -18 < 0\), \(x = 5\) gives a local maximum. 5. Check the endpoints: \(P(0) = -50\), \(P(5) = 50\), and \(P(7) = 6\). Therefore, the absolute maximum occurs at \(x = 5\).

a) \(P(2) = -4\), so the company loses \(\$4{,}000\). b) The maximum profit occurs when the company produces \(5\) tons.

- Profit equals revenue minus cost. - In this model, what quantity does \(x = 1\) represent? - Use derivatives to locate the maximum of the profit function. - Compare interior critical points with the endpoints.

1. Form the profit function: \(P(x) = 25x - (x^3 - 9x^2 + 40x + 10) = -x^3 + 9x^2 - 15x - 10\). 2. Producing \(100\) components means \(x = 1\). Then \(P(1) = -1 + 9 - 15 - 10 = -17\), so the company has a loss of \(\$17{,}000\). 3. Differentiate: \(P'(x) = -3x^2 + 18x - 15\). Solving \(P'(x) = 0\) gives \(x = 1\) and \(x = 5\). 4. Use \(P''(x) = -6x + 18\). Since \(P''(1) = 12 > 0\), \(x = 1\) is a local minimum. Since \(P''(5) = -12 < 0\), \(x = 5\) is a local maximum. 5. Compare the candidate and endpoint values: \(P(0) = -10\), \(P(5) = 15\), and \(P(8) = -66\). Thus, the absolute maximum is \(15\) thousand dollars at \(x = 5\).

a) The company loses \(\$17{,}000\) when it produces \(100\) components. b) The maximum profit is \(\$15{,}000\), achieved by producing \(500\) components.

- How does the sign of the first derivative determine whether a function is increasing? - Use the discriminant to decide whether the derivative can equal zero. - Express revenue in the same units as cost. - Profit is revenue minus cost, and its maximum must be checked on the entire interval.

1. Differentiate the cost function: \(K'(x) = \frac{1}{10}x^2 - x + 4\). 2. The discriminant of \(\frac{1}{10}x^2 - x + 4 = 0\) is \((-1)^2 - 4(\frac{1}{10})(4) = -\frac{3}{5}\). Because the quadratic has no real zeros and a positive leading coefficient, \(K'(x) > 0\) for every \(x\). Thus, \(K\) is strictly increasing. 3. Revenue, in thousands of dollars, is \(R(x) = 6.4x\). Therefore, profit is \(P(x) = R(x) - K(x) = -\frac{1}{30}x^3 + 0.5x^2 + 2.4x - 20\). 4. Differentiate: \(P'(x) = -0.1x^2 + x + 2.4\). Solving \(P'(x) = 0\) gives \(x = -2\) and \(x = 12\). Only \(x = 12\) lies in the domain. 5. Since \(P''(x) = -0.2x + 1\) and \(P''(12) = -1.4 < 0\), \(x = 12\) gives a local maximum. 6. Compare values: \(P(0) = -20\), \(P(12) = 23.2\), and \(P(20) = -\frac{116}{3}\). Therefore, the maximum profit is \(23.2\) thousand dollars, or \(\$23{,}200\), at \(x = 12\).

a) \(K'(x) = 0.1x^2 - x + 4\) is positive for all \(x\), so \(K\) is strictly increasing. b) The maximum profit is \(\$23{,}200\), achieved by producing \(12\) tons.

- First relate the total amount of fencing to the three fenced sides of the rectangle. - Rewrite the area so that it depends on only one variable. - Determine the smallest and largest possible values of the perpendicular side length. - Use the derivative to find an interior critical point. - Compare the critical-point value with the values at both endpoints.

1. Let \(x\) be the side parallel to the wall and \(y\) be each side perpendicular to the wall. The fencing constraint is \(x+2y=60\), so \(x=60-2y\). Therefore, \(A(y)=xy=(60-2y)y=60y-2y^2\). The side lengths must be nonnegative, so \(0\le y\le 30\). Thus the domain is \([0, 30]\). 2. Differentiate: \(A'(y)=60-4y\). Set the derivative equal to zero: \(60-4y=0\Rightarrow y=15\). Then \(x=60-2(15)=30\). Also, \(A''(y)=-4<0\), so the critical point is a local maximum. 3. The area is \(A(15)=30\cdot 15=450\,\text{m}^2\). At the endpoints, \(A(0)=0\) and \(A(30)=0\). Because \(A\) is continuous on \([0, 30]\) and the critical-point value is greater than both endpoint values, \(450\,\text{m}^2\) is the absolute maximum.

1. \(A(y)=60y-2y^2\), with domain \([0, 30]\). 2. The perpendicular sides should each be \(15\,\text{m}\), and the side parallel to the wall should be \(30\,\text{m}\). 3. The maximum area is \(450\,\text{m}^2\). Since \(A(0)=A(30)=0\), the interior maximum is the absolute maximum on the domain.

- Recall the perimeter formula for a square and the volume formula for a rectangular prism. - Use the given sum to express the height in terms of the base side length. - Apply the power rule when differentiating the volume function. - Check that the resulting dimensions are positive and compare with the endpoint cases.

1. The square base has perimeter \(4x\), so the constraint is \(h+4x=108\). Thus \(h=108-4x\). Since \(V=x^2h\), \(V(x)=x^2(108-4x)=108x^2-4x^3\). The physically meaningful domain is \(0\le x\le 27\). 2. Differentiate: \(V'(x)=216x-12x^2=12x(18-x)\). The interior critical point is \(x=18\). The second derivative is \(V''(x)=216-24x\), and \(V''(18)=-216<0\), so this point gives a local maximum. The corresponding height is \(h=108-4(18)=36\,\text{cm}\). The endpoint volumes are both \(0\), so this local maximum is also the absolute maximum on the domain. 3. \(V(18)=18^2\cdot 36=11{,}664\,\text{cm}^3\).

1. Using \(h=108-4x\) and \(V=x^2h\) gives \(V(x)=108x^2-4x^3\). 2. The volume is maximized when \(x=18\,\text{cm}\) and \(h=36\,\text{cm}\). 3. The maximum volume is \(11{,}664\,\text{cm}^3\).

- Express both the fee and the membership count in terms of the fee change. - Revenue equals price per member times number of members. - A downward-opening quadratic reaches its maximum at its vertex. - Report the fee, membership, and revenue.

1. Let \(x\) be the change in the monthly fee, in dollars. Then the fee is \(p(x)=45+x\), and the number of members is \(n(x)=800-10x\). 2. The revenue function is \(R(x)=(45+x)(800-10x)=-10x^2+350x+36{,}000\). 3. Differentiate: \(R'(x)=-20x+350\). Setting \(R'(x)=0\) gives \(x=17.5\). 4. Since \(R''(x)=-20<0\), this critical point gives the maximum. 5. The optimal fee is \(\$45.00+\$17.50=\$62.50\). The corresponding membership is \(800-10(17.5)=625\). 6. The maximum monthly revenue is \(\$62.50\cdot 625=\$39{,}062.50\).

The revenue is maximized with a monthly fee of \(\$62.50\). The center then has \(625\) members and monthly revenue of \(\$39{,}062.50\).

- Translate the product condition into an equation. - Express one number in terms of the other. - Write the quantity to be minimized as a function of one variable. - Remember that both numbers must be positive.

1. Let the first number be \(x\) and the second number be \(y\), with \(x > 0\) and \(y > 0\). 2. The constraint is \(xy = 36\), so \(y = \frac{36}{x}\). 3. The quantity to minimize is \(S = x + 9y\). Substitute the constraint: \(S(x) = x + \frac{324}{x}\). 4. Differentiate: \(S'(x) = 1 - \frac{324}{x^2}\). Setting \(S'(x) = 0\) gives \(x^2 = 324\), so \(x = 18\) because \(x\) is positive. 5. Since \(S''(x) = \frac{648}{x^3} > 0\) for \(x > 0\), the critical point is a minimum. 6. Then \(y = \frac{36}{18} = 2\).

The first number is \(18\), and the second number is \(2\).

- Write an equation for the sum of the two numbers. - Express the product as a function of one variable. - Determine the allowable interval for the first number. - Compare all critical points and endpoints.

1. Let the first number be \(x\) and the second number be \(y\), where \(x \ge 0\) and \(y \ge 0\). 2. Since \(x + y = 15\), write \(y = 15 - x\). 3. The product is \(P(x) = x^2(15 - x) = 15x^2 - x^3\), with \(0 \le x \le 15\). 4. Differentiate: \(P'(x) = 30x - 3x^2 = 3x(10 - x)\). The critical numbers are \(x = 0\) and \(x = 10\). 5. Evaluate the candidates: \(P(0) = 0\), \(P(10) = 500\), and \(P(15) = 0\). Therefore, the absolute maximum occurs at \(x = 10\). 6. The second number is \(y = 15 - 10 = 5\).

The first number is \(10\), and the second number is \(5\).

- Express the rectangle's area using the coordinates of \(P\). - Write an area function that depends only on \(x\). - How do you find the largest value of a function on a closed interval? - Remember to check the endpoints as well as the critical points.

1. The rectangle has width \(x\) and height \(f(x)=(x-6)^2\), so its area is \(A(x)=x(x-6)^2=x^3-12x^2+36x\). 2. Differentiate: \(A'(x)=3x^2-24x+36=3(x-2)(x-6)\). 3. The critical points in \([0,6]\) are \(x=2\) and \(x=6\). 4. Evaluate the candidates: \(A(0)=0\), \(A(2)=32\), and \(A(6)=0\). Therefore, the absolute maximum occurs at \(x=2\). 5. The corresponding height is \(f(2)=(2-6)^2=16\), so \(P=(2,16)\).

\(P=(2,16)\)

- Use the volume formula for a cylinder to express the height in terms of the radius. - Write the total surface area of a closed cylinder and substitute for the height. - Find a critical point of the one-variable surface-area function. - Verify that the critical point gives a minimum. - Use the critical-point equation to compare the height with the radius.

1. The volume constraint is \(\pi r^2h=15\), so \(h=\frac{15}{\pi r^2}\), with \(r>0\). 2. A closed cylinder has surface area \(S=2\pi r^2+2\pi rh\). Substituting for \(h\) gives \(S(r)=2\pi r^2+2\pi r\left(\frac{15}{\pi r^2}\right)=2\pi r^2+\frac{30}{r}\). 3. Differentiate: \(S'(r)=4\pi r-\frac{30}{r^2}\). Set the derivative equal to zero: \(4\pi r-\frac{30}{r^2}=0\Rightarrow 4\pi r^3=30\Rightarrow r=\sqrt[3]{\frac{7.5}{\pi}}\approx 1.337\,\text{m}\). 4. Since \(S''(r)=4\pi+\frac{60}{r^3}>0\) for \(r>0\), this critical point is a local minimum. Also, \(S(r)\to\infty\) as \(r\to 0^+\) or as \(r\to\infty\), so it is the absolute minimum. 5. The height is \(h=\frac{15}{\pi r^2}\approx 2.673\,\text{m}\). 6. At the critical point, \(4\pi r^3=30\), so \(15=2\pi r^3\). Therefore, \(h=\frac{15}{\pi r^2}=\frac{2\pi r^3}{\pi r^2}=2r\). Thus the height equals the diameter.

a) \(S(r)=2\pi r^2+\frac{30}{r}\), for \(r>0\). b) \(r=\sqrt[3]{\frac{7.5}{\pi}}\approx 1.337\,\text{m}\) and \(h\approx 2.673\,\text{m}\). c) At the minimum, \(h=2r\), so the height equals the diameter.

- Recall the distance formula for two points in the coordinate plane. - Why can you minimize the square of the distance instead of the distance itself? - Keep the domain restriction \(x\ge -5\) in mind. - Verify that the critical point gives a minimum.

1. The squared distance from \(Q\) to \(P\) is \(D(x)=(x-3)^2+(\sqrt{x+5})^2=x^2-5x+14\). 2. Minimizing the distance is equivalent to minimizing \(D(x)\), because the square-root function is increasing. 3. Differentiate: \(D'(x)=2x-5\). Setting \(D'(x)=0\) gives \(x=\frac{5}{2}\). 4. Since \(D(x)\) is an upward-opening quadratic, this critical point gives its global minimum on \([-5,\infty)\). 5. The minimum distance is \(\sqrt{D\left(\frac{5}{2}\right)}=\sqrt{\frac{31}{4}}=\frac{\sqrt{31}}{2}\approx 2.78\).

The distance is minimized at \(x=\frac{5}{2}\), and the minimum distance is \(\frac{\sqrt{31}}{2}\approx 2.78\).

- Identify which dimensions determine the rectangle’s area and which segments make up the outside perimeter. - Use the perimeter constraint to express one dimension in terms of the other. - Remember that the semicircle is attached to only one short side; the opposite short side remains part of the perimeter. - Maximize the resulting one-variable area function.

1. The area of the rectangle is \(A=bh\). Because the semicircle has diameter \(b\), its arc length is \(\frac{\pi b}{2}\). The perimeter constraint is \(b+2h+\frac{\pi b}{2}=200\). 2. Solve for \(h\): \(h=100-\left(\frac{1}{2}+\frac{\pi}{4}\right)b\). Therefore, \(A(b)=100b-\left(\frac{1}{2}+\frac{\pi}{4}\right)b^2\). 3. Differentiate: \(A'(b)=100-\left(1+\frac{\pi}{2}\right)b\). Setting the derivative equal to zero gives \(b=\frac{100}{1+\frac{\pi}{2}}=\frac{200}{2+\pi}\approx 38.90\,\text{m}\). 4. Substitute into the constraint: \(h=100-\frac{200}{2+\pi}\cdot\frac{2+\pi}{4}=50\,\text{m}\). Since \(A''(b)=-\left(1+\frac{\pi}{2}\right)<0\), the area function is concave down, so these dimensions give the absolute maximum on the physical domain.

The width should be \(b=\frac{200}{2+\pi}\approx 38.90\,\text{m}\), and the length should be \(h=50\,\text{m}\).

- Use the volume relationship among the three dimensions of a rectangular prism. - Write the surface area of a closed rectangular prism in terms of its edge lengths. - Use the length condition and the volume constraint to reduce the problem to one variable. - Find and verify the minimum of the resulting surface-area function.

1. Let the width be \(b\), the length be \(2b\), and the height be \(h\). The volume constraint is \(2b^2h=72\), so \(h=\frac{36}{b^2}\). 2. The surface area is \(S=2(2b\cdot b)+2(2b\cdot h)+2(b\cdot h)=4b^2+6bh\). Substituting for \(h\) gives \(S(b)=4b^2+\frac{216}{b}\), for \(b>0\). 3. Differentiate: \(S'(b)=8b-\frac{216}{b^2}\). Setting \(S'(b)=0\) gives \(8b^3=216\Rightarrow b^3=27\Rightarrow b=3\,\text{ft}\). 4. Since \(S''(b)=8+\frac{432}{b^3}>0\) for \(b>0\), the critical point is a local minimum. Because \(S(b)\to\infty\) as \(b\to 0^+\) or as \(b\to\infty\), it is the absolute minimum. 5. The length is \(2b=6\,\text{ft}\), and the height is \(h=\frac{36}{3^2}=4\,\text{ft}\).

The width is \(3\,\text{ft}\), the length is \(6\,\text{ft}\), and the height is \(4\,\text{ft}\).

- Write the area of the right triangle as one-half times base times height. - How is the height related to \(f(u)\)? - Use the quotient rule to differentiate the area function. - What sign change in the derivative identifies a maximum? - Check that the critical value is in the domain.

1. The triangle has base \(u\) and height \(f(u)\), so \(A(u)=\frac{1}{2}u f(u)=\frac{5\ln(u)}{u}\). 2. Differentiate using the quotient rule: \(A'(u)=\frac{5-5\ln(u)}{u^2}\). 3. Set the derivative equal to zero: \(5-5\ln(u)=0\), so \(\ln(u)=1\) and \(u=e\). 4. For \(1\le u<e\), \(A'(u)>0\), and for \(u>e\), \(A'(u)<0\). Therefore, \(A\) increases and then decreases at \(u=e\). 5. Also, \(A(1)=0\) and \(A(u)\to 0\) as \(u\to\infty\), so \(u=e\) gives the global maximum.

The triangle has maximum area when \(u=e\).

- Express the rectangle's side lengths using the coordinates of \(Q\). - Use both the product rule and the chain rule to differentiate the area function. - The exponential factor is never zero, so which factor determines the critical point? - Verify the maximum using a derivative sign change or the second derivative.

1. The rectangle has width \(u\) and height \(g(u)\), so \(A(u)=12u e^{-0.25u^2}\). 2. Differentiate using the product and chain rules: \(A'(u)=(12-6u^2)e^{-0.25u^2}\). 3. Since the exponential factor is positive, \(A'(u)=0\) when \(12-6u^2=0\). Because \(u\ge 0\), \(u=\sqrt{2}\). 4. The derivative is positive before \(\sqrt{2}\) and negative after \(\sqrt{2}\), so this critical point gives the maximum. 5. The maximum area is \(A(\sqrt{2})=12\sqrt{2}e^{-1/2}\approx 10.29\).

The area is maximized at \(u=\sqrt{2}\), and the maximum area is \(12\sqrt{2}e^{-1/2}\approx 10.29\).

- Substitute the given time into the function. - Use the product and chain rules to differentiate the time-dependent term. - Find where the derivative changes from positive to negative. - Exponential decay eventually dominates polynomial growth.

1. Substitute \(t=0\): \(k(0)=0.8\,\text{mg/L}\). 2. Differentiate using the product and chain rules: \(k'(t)=(12-6t)e^{-0.5t}\). Since the exponential factor is positive, the derivative is zero at \(t=2\), positive before \(2\), and negative after \(2\). Therefore, the maximum occurs after \(2\) hours. 3. The maximum concentration is \(k(2)=0.8+24e^{-1}\approx9.63\,\text{mg/L}\). 4. After \(12\) hours, \(k(12)=0.8+144e^{-6}\approx1.16\,\text{mg/L}\). 5. Since \(te^{-0.5t}\to0\), \(\lim_{t\to\infty}k(t)=0.8\,\text{mg/L}\). In the model, the concentration approaches a baseline level of \(0.8\,\text{mg/L}\).

a) \(0.8\,\text{mg/L}\). b) After \(2\) hours; maximum concentration \(0.8+\frac{24}{e}\approx9.63\,\text{mg/L}\). c) Approximately \(1.16\,\text{mg/L}\). d) \(\lim_{t\to\infty}k(t)=0.8\,\text{mg/L}\); the model approaches that baseline concentration.

- A maximum can occur where the first derivative changes from positive to negative. - Factor the derivative before solving for critical times. - Keep the exponential factor when differentiating. - Exponential decay eventually dominates a polynomial factor.

1. Substitute \(t=0\): \(f(0)=15\,^{\circ}\text{C}\). 2. Differentiate: \(f'(t)=(0.4t-0.02t^2)e^{-0.1t}=t(0.4-0.02t)e^{-0.1t}\). 3. For \(t>0\), the derivative is zero at \(t=20\), positive before \(20\), and negative after \(20\). Therefore, the maximum occurs after \(20\) seconds. 4. The maximum temperature is \(f(20)=15+80e^{-2}\approx25.83\,^{\circ}\text{C}\). 5. After \(100\) seconds, \(f(100)=15+2000e^{-10}\approx15.09\,^{\circ}\text{C}\). 6. Since \(t^2e^{-0.1t}\to0\), \(\lim_{t\to\infty}f(t)=15\,^{\circ}\text{C}\). In the model, the component temperature approaches \(15\,^{\circ}\text{C}\).

a) \(15\,^{\circ}\text{C}\). b) After \(20\) seconds; approximately \(25.83\,^{\circ}\text{C}\). c) Approximately \(15.09\,^{\circ}\text{C}\). d) \(\lim_{t\to\infty}f(t)=15\,^{\circ}\text{C}\); the modeled temperature approaches \(15\,^{\circ}\text{C}\).

- Relate the volume of the contents to the radius and fill height. - Add \(2\,\text{cm}\) to the fill height to obtain the total can height. - Include the curved side, top, and bottom in the surface area. - Substitute for the height, then set the derivative of the one-variable function equal to zero.

1. Since \(800\,\text{mL}=800\,\text{cm}^3\), the fill-height constraint is \(\pi r^2h_{\text{fill}}=800\), so \(h_{\text{fill}}=\frac{800}{\pi r^2}\). 2. The total height is \(h=\frac{800}{\pi r^2}+2\). 3. A closed cylinder has surface area \(S=2\pi r^2+2\pi rh\). Therefore, \(S(r)=2\pi r^2+2\pi r\left(\frac{800}{\pi r^2}+2\right)=2\pi r^2+\frac{1600}{r}+4\pi r\), for \(r>0\). 4. Differentiate: \(S'(r)=4\pi r-\frac{1600}{r^2}+4\pi\). The critical-point equation is \(\pi r^3+\pi r^2-400=0\). Solving numerically gives \(r\approx 4.719\,\text{cm}\). 5. Since \(S''(r)=4\pi+\frac{3200}{r^3}>0\) for \(r>0\), this critical point gives a minimum. The total height is \(h=\frac{800}{\pi(4.718577\ldots)^2}+2\approx 13.437\,\text{cm}\).

a) \(S(r)=2\pi r^2+\frac{1600}{r}+4\pi r\), for \(r>0\). b) The minimizing dimensions are \(r\approx 4.719\,\text{cm}\) and \(h\approx 13.437\,\text{cm}\).

- Find the horizontal base length and vertical height of the triangle. - Use \(A=\frac{1}{2}bh\). - Express the point coordinates and height in terms of \(k\). - Differentiate the area function to locate its largest value. - Check the derivative sign and the behavior near the ends of the domain.

1. For \(k=1\), \(f(1)=\frac{1}{5}\). The base \(P_1Q_1\) has length \(2\), and the height from \(R\) is \(1-\frac{1}{5}=\frac{4}{5}\). Thus the area is \(\frac{1}{2}\cdot 2\cdot\frac{4}{5}=\frac{4}{5}\). 2. In general, the base has length \(2k\), and the height is \(1-f(k)=1-\frac{k^2}{k^2+4}=\frac{4}{k^2+4}\). 3. Therefore, \(A(k)=\frac{1}{2}(2k)\left(\frac{4}{k^2+4}\right)=\frac{4k}{k^2+4}\). 4. Differentiate: \(A'(k)=\frac{16-4k^2}{(k^2+4)^2}\). 5. For \(k>0\), the only critical point is \(k=2\). The derivative is positive for \(0<k<2\) and negative for \(k>2\), so this is a maximum. 6. Since \(A(k)\to 0\) as \(k\to 0^+\) and as \(k\to\infty\), the maximum is global. Its value is \(A(2)=1\).

a) The area for \(k=1\) is \(\frac{4}{5}=0.8\), and \(A(k)=\frac{4k}{k^2+4}\). b) The maximum occurs at \(k=2\), and the maximum area is \(1\).

- First use the log’s diameter to relate the width and height of the inscribed rectangle. - The area and load-capacity objectives lead to different one-variable functions. - Find the maximizing dimensions separately for each objective. - In the percent comparison, the proportionality constant \(k\) cancels.

1. The log has diameter \(60\,\text{cm}\). Because the rectangular cross section is inscribed in the circle, \(b^2+h^2=60^2=3600\). 2. For part a, the area is \(A(b)=b\sqrt{3600-b^2}\). Differentiating gives \(A'(b)=\frac{3600-2b^2}{\sqrt{3600-b^2}}\). The interior critical point satisfies \(b^2=1800\), so \(b=h=\sqrt{1800}=30\sqrt{2}\approx 42.43\,\text{cm}\). The area is \(0\) at the endpoints of the physical domain, so this critical point gives the absolute maximum. 3. For part b, \(T(b)=kb(3600-b^2)=k(3600b-b^3)\). Then \(T'(b)=k(3600-3b^2)\). The critical point satisfies \(b^2=1200\), so \(b=20\sqrt{3}\approx 34.64\,\text{cm}\) and \(h=\sqrt{3600-1200}=20\sqrt{6}\approx 48.99\,\text{cm}\). Again, the endpoint values are \(0\), so this is the absolute maximum. 4. For part c, the common factor \(k\) cancels. The two proportional load-capacity values are \(T_a=(30\sqrt{2})(1800)=54{,}000\sqrt{2}\) and \(T_b=(20\sqrt{3})(2400)=48{,}000\sqrt{3}\). Therefore, \(\frac{T_b-T_a}{T_a}\cdot 100\%=\left(\frac{8\sqrt{6}}{18}-1\right)100\%\approx 8.87\%\).

a) \(b=h=30\sqrt{2}\approx 42.43\,\text{cm}\). b) \(b=20\sqrt{3}\approx 34.64\,\text{cm}\) and \(h=20\sqrt{6}\approx 48.99\,\text{cm}\). c) The load capacity is about \(8.87\%\) greater.

- Express the triangle's width in terms of its height \(k\). - Use the triangle area formula. - Use the parabola's symmetry to find the full base length. - Confirm that the maximizing value lies in \(0<k<12\).

1. The intersections satisfy \(k=12-x^2\), so \(x=\pm\sqrt{12-k}\). 2. The horizontal base \(PQ\) has length \(2\sqrt{12-k}\), and the height from the origin to \(y=k\) is \(k\). 3. Therefore, \(A(k)=\frac{1}{2}\left(2\sqrt{12-k}\right)k=k\sqrt{12-k}\), for \(0<k<12\). 4. Differentiate: \(A'(k)=\frac{24-3k}{2\sqrt{12-k}}\). 5. The derivative is zero at \(k=8\), positive before \(8\), and negative after \(8\). Also, \(A(k)\to 0\) as \(k\to 0^+\) and as \(k\to 12^-\), so \(k=8\) gives the global maximum. 6. The maximum area is \(A(8)=8\sqrt{4}=16\).

The area is maximized when \(k=8\), and the maximum area is \(16\) square units.

- Find the equation of the line through \(B\) and \(C\). - Express the horizontal length \(DE\) in terms of \(k\). - The height of triangle \(DEM\) relative to base \(DE\) is the distance from \(M\) to the line \(y=k\). - Consider whether the x-coordinate of \(M\) affects that distance.

1. The line through \(B(12, 0)\) and \(C(0, 6)\) has slope \(-\frac{1}{2}\), so the hypotenuse has equation \(y=-\frac{1}{2}x+6\). 2. Since \(D\) lies on the y-axis, \(D=(0, k)\). At point \(E\), \(k=-\frac{1}{2}x+6\), so \(x=12-2k\) and \(E=(12-2k, k)\). 3. The horizontal base \(DE\) has length \(12-2k\). The perpendicular distance from \(M\) to the line \(y=k\) is \(k\). Thus \(A(k)=\frac{1}{2}(12-2k)k=6k-k^2\), for \(0<k<6\). 4. Differentiate: \(A'(k)=6-2k\). Setting the derivative equal to zero gives \(k=3\). Since \(A''(k)=-2<0\), the function is concave down and \(k=3\) gives the absolute maximum.

Segment \(DE\) should be drawn at \(k=3\).

- Use similar triangles in a cross section through the cone’s axis. - Express the cylinder height in terms of its radius. - Maximize the resulting one-variable cylinder-volume function. - Divide the maximum cylinder volume by the cone volume.

1. Similar triangles in an axial cross section give \(\frac{h}{H}=\frac{R-r}{R}\), so \(h=H\left(1-\frac{r}{R}\right)\). 2. The cylinder volume is \(V(r)=\pi r^2H\left(1-\frac{r}{R}\right)=\pi H\left(r^2-\frac{r^3}{R}\right)\), for \(0\le r\le R\). 3. Differentiate: \(V'(r)=\pi H\left(2r-\frac{3r^2}{R}\right)=\pi Hr\left(2-\frac{3r}{R}\right)\). The interior critical point is \(r=\frac{2R}{3}\). Since the volume is \(0\) at \(r=0\) and \(r=R\), this critical point gives the absolute maximum. 4. The corresponding height is \(h=H\left(1-\frac{2}{3}\right)=\frac{H}{3}\). 5. The maximum cylinder volume is \(V_{\text{cyl}}=\pi\left(\frac{2R}{3}\right)^2\left(\frac{H}{3}\right)=\frac{4}{27}\pi R^2H\). The cone volume is \(V_{\text{cone}}=\frac{1}{3}\pi R^2H\), so \(\frac{V_{\text{cyl}}}{V_{\text{cone}}}=\frac{4}{9}\).

a) \(r=\frac{2R}{3}\) and \(h=\frac{H}{3}\). b) The cylinder occupies at most \(\frac{4}{9}\), or about \(44.4\%\), of the cone’s volume.

- Relate the radius of the rotated solid to the altitude from the right angle to the hypotenuse. - Add the volumes of the two cones, noting that their heights sum to \(12\,\text{cm}\). - Use the Pythagorean theorem to eliminate one leg. - A substitution such as \(x=a^2\) simplifies the objective function.

1. The legs satisfy \(a^2+b^2=12^2=144\). The radius of the double cone is the altitude to the hypotenuse: \(r=\frac{ab}{12}\). 2. The two cone heights add to the hypotenuse length, so the total volume is \(V=\frac{1}{3}\pi r^2(12)=\frac{\pi}{36}a^2b^2\). Using \(b^2=144-a^2\), \(V(a)=\frac{\pi}{36}a^2(144-a^2)\). 3. Let \(x=a^2\), where \(0\le x\le 144\). Then \(V(x)=\frac{\pi}{36}(144x-x^2)\). Differentiate: \(V'(x)=\frac{\pi}{36}(144-2x)\). The critical point is \(x=72\). Since this quadratic is concave down, it gives the absolute maximum. 4. Therefore, \(a=b=\sqrt{72}=6\sqrt{2}\approx 8.49\,\text{cm}\). 5. The maximum volume is \(V=\frac{\pi}{36}(72)(72)=144\pi\approx 452.39\,\text{cm}^3\).

a) \(a=b=6\sqrt{2}\approx 8.49\,\text{cm}\). b) The maximum volume is \(144\pi\approx 452.39\,\text{cm}^3\).

- Express the box’s length, width, and height in terms of the corner-square size \(x\). - Determine the physically meaningful domain for \(x\). - Differentiate the volume function and discard critical points outside the domain. - Compare the remaining critical-point value with the endpoint values.

1. The box dimensions are \(48-2x\), \(30-2x\), and \(x\). Thus \(V(x)=x(48-2x)(30-2x)\). The physical domain is \(0\le x\le 15\). 2. Expand: \(V(x)=4x^3-156x^2+1440x\). Differentiate: \(V'(x)=12x^2-312x+1440=12(x^2-26x+120)\). The critical points are \(x=6\) and \(x=20\). Only \(x=6\) lies in the physical domain. 3. The endpoint volumes are \(V(0)=V(15)=0\), while \(V(6)=6(36)(18)=3888\,\text{cm}^3\). Therefore, \(x=6\,\text{cm}\) gives the absolute maximum.

The squares should have side length \(x=6\,\text{cm}\). The maximum volume is \(3888\,\text{cm}^3\).

- Profit equals revenue minus cost. - Use the first derivative to find candidates for maximum profit. - Use the second derivative or a sign chart to verify the maximum. - A loss occurs where the profit function is negative. - Approximate the break-even points numerically.

1. Revenue is \(R(x)=500x\), so profit is \(P(x)=R(x)-C(x)=-x^3+30x^2-1000\). 2. Differentiate: \(P'(x)=-3x^2+60x=-3x(x-20)\). The critical values for \(x\ge 0\) are \(x=0\) and \(x=20\). 3. Since \(P''(20)=-60<0\), and \(P\) increases on \((0,20)\) and decreases for \(x>20\), the global maximum occurs at \(x=20\). 4. The maximum profit is \(P(20)=-8000+12{,}000-1000=3000\), or \(\$3000\). 5. Numerically, the nonnegative break-even points are \(x\approx 6.53\) and \(x\approx 28.79\). Thus the continuous model gives a loss for \(0\le x<6.53\) and for \(x>28.79\). 6. For whole-number production levels, the company loses money at \(x=0,1,\ldots,6\) and at \(x\ge 29\).

a) \(P(x)=-x^3+30x^2-1000\) b) The maximum occurs at \(x=20\) units, with a profit of \(\$3000\). c) In the continuous model, profit is negative for \(0\le x<6.53\) and \(x>28.79\). For whole units, there is a loss at \(0\) through \(6\) units and at \(29\) or more units.

- Which function are you minimizing? - How do derivatives locate a minimum? - How is marginal cost related to total cost? - Check both the necessary and sufficient conditions.

1. Differentiate to obtain the marginal cost: \(K'(x)=\frac{1}{100}x^2-x+40\). 2. Differentiate the marginal-cost function: \(K''(x)=\frac{1}{50}x-1\). 3. Solve \(K''(x)=0\): \(\frac{1}{50}x-1=0\), so \(x=50\). 4. Since \(K'''(x)=\frac{1}{50}>0\), the marginal-cost function has a minimum at \(x=50\). 5. Evaluate \(K'(50)=\frac{1}{100}\cdot50^2-50+40=15\).

The marginal cost is minimized at \(x=50\) units, and the minimum marginal cost is \(\$15\) per unit.

- Profit equals revenue minus cost. - Use the first- and second-derivative tests to maximize profit. - Discard production levels that are not feasible. - A negative profit value represents a loss.

1. Revenue is \(E(x)=600x\). Therefore, \(G(x)=E(x)-K(x)=-0.05x^3+6x^2+135x-4700\). 2. The derivative is \(G'(x)=-0.15x^2+12x+135\). Solving \(G'(x)=0\) gives \(x=-10\) and \(x=90\). Only \(x=90\) is a feasible production level. Since \(G''(90)=-15<0\), profit is maximized there. The maximum profit is \(G(90)=\$19{,}600\). 3. \(G(10)=-2800\), so producing \(10\,\text{kg}\) results in a loss of \(\$2800\).

a) \(G(x)=-0.05x^3+6x^2+135x-4700\) b) \(90\,\text{kg}\); maximum profit \(\$19{,}600\) c) A loss of \(\$2800\)

- Marginal cost is the first derivative of total cost. - Minimize the marginal-cost function. - Use the sign of the second derivative of total cost and restrict the result to nonnegative production levels.

1. The marginal-cost function is \(K'(x)=0.6x^2-24x+300\). 2. Differentiate again: \(K''(x)=1.2x-24\). Setting this equal to zero gives \(x=20\). Since \(K'''(x)=1.2>0\), marginal cost has a minimum there. The minimum value is \(K'(20)=\$60\,\text{per unit}\). 3. Since the minimum marginal cost is \(\$60\,\text{per unit}>0\), total cost is increasing for all feasible production levels. It increases at a decreasing rate where marginal cost is decreasing, which means \(K''(x)<0\). Solving \(1.2x-24<0\) gives \(x<20\). For feasible production levels, this is \(0\le x<20\).

a) \(K'(x)=0.6x^2-24x+300\) b) At \(x=20\) units; \(\$60\,\text{per unit}\) c) \(0\le x<20\)

- Fixed cost is the cost when production is zero. - Marginal cost is the first derivative of total cost. - Minimize the marginal-cost function. - Show that the first derivative is always positive.

1. The fixed cost is the cost at zero production: \(K(0)=\$60\). 2. The marginal-cost function is \(K'(x)=x^2-8x+21\). Its derivative is \(K''(x)=2x-8\), which is zero at \(x=4\). Since \(K'''(x)=2>0\), marginal cost is minimized at \(x=4\). 3. Complete the square: \(K'(x)=x^2-8x+21=(x-4)^2+5>0\) for every real \(x\). Therefore, \(K\) is strictly increasing on \([0, \infty)\).

a) \(\$60\) b) \(x=4\) production units c) \(K'(x)=(x-4)^2+5>0\), so \(K\) is strictly increasing for all \(x\ge0\).

- No fertilizer corresponds to \(x=0\). - Find the absolute maximum on the stated interval. - Marginal yield is the first derivative. - Maximize the first derivative to find the greatest marginal increase.

1. With no fertilizer, \(x=0\), so \(f(0)=45\) bushels per acre. 2. The derivative is \(f'(x)=-0.00015x^2+0.012x+0.3\). Solving \(f'(x)=0\) gives \(x=-20\) and \(x=100\), so the relevant critical value is \(x=100\). Since \(f''(100)=-0.018<0\), this is a local maximum. Checking the endpoints confirms that the maximum modeled yield is \(f(100)=85\) bushels per acre. 3. The marginal increase in yield is \(f'(x)\). It is greatest where \(f''(x)=-0.0003x+0.012=0\), which gives \(x=40\) pounds per acre.

a) \(45\) bushels per acre b) \(100\) pounds per acre; \(85\) bushels per acre c) \(40\) pounds per acre

- Identify the fixed surface area and the volume to be maximized. - Write the surface area of a closed rectangular box with a square base. - Use the surface-area constraint to eliminate the height. - Find and verify the maximum of the one-variable volume function.

1. Let \(a\) be the side length of the square base and \(h\) be the height. The volume is \(V=a^2h\), and the surface-area constraint is \(2a^2+4ah=150\). 2. Solve for \(h\): \(h=\frac{150-2a^2}{4a}=\frac{37.5}{a}-\frac{a}{2}\). Positive dimensions require \(0<a<\sqrt{75}\). 3. Substitute into the volume: \(V(a)=37.5a-\frac{1}{2}a^3\). Differentiate: \(V'(a)=37.5-1.5a^2\). The positive critical point satisfies \(a^2=25\), so \(a=5\,\text{cm}\). 4. Since \(V''(a)=-3a<0\) for \(a>0\), the critical point is a local maximum. The volume approaches \(0\) at both ends of the physical domain, so it is the absolute maximum. 5. The height is \(h=\frac{150-2(5)^2}{4(5)}=5\,\text{cm}\).

The base side length is \(5\,\text{cm}\), and the height is \(5\,\text{cm}\). The maximizing box is a cube.

- Differentiate the product of sine and cosine. - Set the first derivative equal to zero. - Use the second derivative to verify a maximum. - Convert radians to degrees using \(180^\circ=\pi\) radians.

1. Differentiate using the product rule: \(d'(\alpha)=120(\cos^2\alpha-\sin^2\alpha)\). 2. Set the derivative equal to zero: \(\cos^2\alpha-\sin^2\alpha=0\). Since \(\alpha\) is in the first quadrant, \(\tan\alpha=1\), so \(\alpha=\frac{\pi}{4}\). 3. Differentiate again: \(d''(\alpha)=-480\sin\alpha\cos\alpha\). At \(\alpha=\frac{\pi}{4}\), \(d''\left(\frac{\pi}{4}\right)=-240<0\), so the range is maximized there. 4. Convert to degrees: \(\frac{\pi}{4}\cdot\frac{180^\circ}{\pi}=45^\circ\).

\(\alpha=\frac{\pi}{4}=45^\circ\)

- Write the area as a function of \(x\). - Use the derivative to find critical points inside the interval. - Compare every critical value with both endpoint values. - An absolute maximum can occur at more than one input.

1. The area function is \(A(x)=x((x-2)^2+1)=x^3-4x^2+5x\). 2. Differentiate: \(A'(x)=3x^2-8x+5=(3x-5)(x-1)\). 3. The critical points are \(x=1\) and \(x=\frac{5}{3}\). 4. Evaluate the area at the critical points and endpoints: \(A(0)=0\), \(A(1)=2\), \(A\left(\frac{5}{3}\right)=\frac{50}{27}\), and \(A(2)=2\). 5. The largest value is \(2\), attained at both \(x=1\) and \(x=2\).

The maximum area is \(2\) square units, attained at \(x=1\) and \(x=2\).

- Include one square base and four rectangular sides in the material area. - Use the fixed volume to express the height in terms of the base side length. - Minimize the resulting one-variable area function. - Use the critical-point equation to simplify the ratio \(a:h\).

1. The material area is \(A=a^2+4ah\), and the volume constraint is \(V=a^2h\). Thus \(h=\frac{V}{a^2}\). 2. Substitute into the area formula: \(A(a)=a^2+\frac{4V}{a}\), for \(a>0\). 3. Differentiate: \(A'(a)=2a-\frac{4V}{a^2}\). Setting the derivative equal to zero gives \(2a^3=4V\Rightarrow a^3=2V\), so \(a=\sqrt[3]{2V}\). 4. Since \(A''(a)=2+\frac{8V}{a^3}>0\) for \(a>0\), the critical point gives a minimum. The corresponding height is \(h=\frac{V}{a^2}=\frac{a}{2}\). Therefore, \(a:h=2:1\).

The material is minimized when \(a=2h\), so the optimal ratio is \(a:h=2:1\).

- Write a formula for the total enclosed area. - Account for all three perpendicular fence segments in the fencing constraint. - Use the constraint to eliminate one variable. - Find the vertex of the resulting area function and verify that it is a maximum.

1. The area is \(A=xy\), and the fencing constraint is \(3x+y=120\). Thus \(y=120-3x\). 2. Substitute into the area: \(A(x)=x(120-3x)=120x-3x^2\), for \(0\le x\le 40\). 3. Differentiate: \(A'(x)=120-6x\). The critical point is \(x=20\,\text{m}\). Since \(A''(x)=-6<0\), the area function is concave down, so this point gives the absolute maximum. 4. The corresponding length is \(y=120-3(20)=60\,\text{m}\).

The maximum area occurs when \(x=20\,\text{m}\) and \(y=60\,\text{m}\).

- Distinguish between the printed dimensions and the full poster dimensions. - Use the fixed printed area to express one printed dimension in terms of the other. - Write the full paper area, including all margins. - Minimize the resulting one-variable function.

1. Let \(x\) and \(y\) be the width and height of the printed region. Then \(xy=384\), while the full poster dimensions are \(x+4\) and \(y+6\). 2. The total area is \(A=(x+4)(y+6)\). Using \(y=\frac{384}{x}\), \(A(x)=(x+4)\left(\frac{384}{x}+6\right)=6x+\frac{1536}{x}+408\), for \(x>0\). 3. Differentiate: \(A'(x)=6-\frac{1536}{x^2}\). The critical point satisfies \(x^2=256\), so \(x=16\,\text{cm}\). 4. Since \(A''(x)=\frac{3072}{x^3}>0\) for \(x>0\), the critical point gives a minimum. The printed height is \(y=\frac{384}{16}=24\,\text{cm}\). Therefore, the overall dimensions are \(16+4=20\,\text{cm}\) by \(24+6=30\,\text{cm}\).

The poster should be \(20\,\text{cm}\) wide and \(30\,\text{cm}\) high.

- Add the areas of the rectangle and equilateral triangle. - Include only exterior edges in the perimeter. - Use the perimeter constraint to express \(h\) in terms of \(b\). - Maximize the resulting quadratic area function.

1. The area is \(A=bh+\frac{\sqrt{3}}{4}b^2\). The shared edge between the rectangle and triangle is internal, so the outside perimeter is \(3b+2h=300\). Thus \(h=150-\frac{3}{2}b\). 2. Substitute into the area: \(A(b)=150b-\frac{6-\sqrt{3}}{4}b^2\). 3. Differentiate: \(A'(b)=150-\frac{6-\sqrt{3}}{2}b\). The critical point is \(b=\frac{300}{6-\sqrt{3}}\approx 70.29\,\text{cm}\). Because the coefficient of \(b^2\) is negative, the area function is concave down, so this point gives the absolute maximum. 4. The rectangular height is \(h=150-\frac{3}{2}\left(\frac{300}{6-\sqrt{3}}\right)=\frac{450-150\sqrt{3}}{6-\sqrt{3}}\approx 44.56\,\text{cm}\).

The window area is maximized when \(b=\frac{300}{6-\sqrt{3}}\approx 70.29\,\text{cm}\) and \(h=\frac{450-150\sqrt{3}}{6-\sqrt{3}}\approx 44.56\,\text{cm}\).

- Use the equation of the diagonal cut to relate \(x\) and \(y\). - Write the rectangle’s area as a function of \(x\). - Find the critical point within \([6, 16]\). - Compare the critical-point area with both endpoint areas.

1. Along the cut edge, \(y=-0.5x+13\). The area of a rectangle whose upper-right corner lies on this edge is \(A(x)=x(-0.5x+13)=-0.5x^2+13x\), for \(6\le x\le 16\). 2. Differentiate: \(A'(x)=-x+13\). The critical point is \(x=13\). Since \(A''(x)=-1<0\), it gives a local maximum. 3. The corresponding height is \(y=-0.5(13)+13=6.5\). The endpoint areas are \(A(6)=60\) and \(A(16)=80\), while \(A(13)=84.5\). Therefore, the absolute maximum occurs at \(x=13\,\text{in}\) and \(y=6.5\,\text{in}\).

The maximizing side lengths are \(x=13\,\text{in}\) and \(y=6.5\,\text{in}\).

- Weight the base and top areas by twice the side-wall cost rate. - Use the fixed volume to express the height in terms of the base side length. - Ignore any positive constant cost factor that does not affect the location of the minimum. - Minimize and verify the resulting one-variable cost function.

1. Let the side-wall material cost \(c\) per square foot. The base and top have total area \(2x^2\) and cost rate \(2c\), while the four walls have total area \(4xh\) and cost rate \(c\). Thus \(C=4cx^2+4cxh\). The positive constant \(c\) does not affect the minimizing dimensions, so minimize \(F=4x^2+4xh\). 2. The volume constraint is \(x^2h=54\), so \(h=\frac{54}{x^2}\). 3. Substitute: \(F(x)=4x^2+\frac{216}{x}\), for \(x>0\). Differentiate: \(F'(x)=8x-\frac{216}{x^2}\). The critical point satisfies \(8x^3=216\), so \(x=3\,\text{ft}\). 4. Since \(F''(x)=8+\frac{432}{x^3}>0\) for \(x>0\), the critical point gives a minimum. The height is \(h=\frac{54}{3^2}=6\,\text{ft}\).

The total material cost is minimized when \(x=3\,\text{ft}\) and \(h=6\,\text{ft}\).

- Use the side relationships in an isosceles right triangle to write the perimeter constraint. - Convert each absorption rate to a transmission rate. - Weight each region’s area by its transmission rate. - Substitute the perimeter constraint and maximize the resulting quadratic function.

1. If the triangle’s hypotenuse is \(x\), each leg has length \(\frac{x}{\sqrt{2}}\). The perimeter constraint is \(x+2y+\sqrt{2}x=8\), so \(y=4-\frac{1+\sqrt{2}}{2}x\). 2. The rectangle area is \(xy\). The triangle area is \(\frac{1}{2}\left(\frac{x}{\sqrt{2}}\right)^2=\frac{x^2}{4}\). The transmitted-light objective is proportional to \(L=0.9xy+0.6\left(\frac{x^2}{4}\right)\). 3. Substitute for \(y\): \(L(x)=3.6x-(0.3+0.45\sqrt{2})x^2\). Differentiate: \(L'(x)=3.6-(0.6+0.9\sqrt{2})x\). The critical point is \(x=\frac{3.6}{0.6+0.9\sqrt{2}}\approx 1.922\,\text{m}\). 4. Because the quadratic coefficient is negative, this point gives the absolute maximum. The corresponding rectangle height is \(y=4-\frac{1+\sqrt{2}}{2}(1.922263\ldots)\approx 1.680\,\text{m}\).

The transmitted light is maximized when \(x\approx 1.922\,\text{m}\) and \(y\approx 1.680\,\text{m}\).

- Use the fixed volume to express the height in terms of the base side length. - Include one base and four side walls in the cost function. - Apply the two different material rates to the correct areas. - Minimize the resulting one-variable cost function.

1. The volume constraint is \(x^2h=6\), so \(h=\frac{6}{x^2}\). 2. The base costs \(12x^2\), and the four side walls cost \(8(4xh)=32xh\). Thus \(C(x, h)=12x^2+32xh\). Substituting for \(h\) gives \(C(x)=12x^2+\frac{192}{x}\), for \(x>0\). 3. Differentiate: \(C'(x)=24x-\frac{192}{x^2}\). The critical point satisfies \(24x^3=192\Rightarrow x^3=8\), so \(x=2\,\text{m}\). 4. Since \(C''(x)=24+\frac{384}{x^3}>0\) for \(x>0\), the critical point gives a minimum. The height is \(h=\frac{6}{2^2}=1.5\,\text{m}\).

The total cost is minimized when \(x=2\,\text{m}\) and \(h=1.5\,\text{m}\).

- Use similar triangles in an axial cross section to relate \(r\) and \(h\). - The lateral area of a cylinder is circumference times height. - Maximize the resulting quadratic function of \(h\). - Use the cone’s slant height to find its lateral surface area.

1. Similar triangles give \(\frac{r}{R}=\frac{H-h}{H}\), so \(r=R\left(1-\frac{h}{H}\right)\). Therefore, \(M(h)=2\pi rh=2\pi R\left(h-\frac{h^2}{H}\right)\), for \(0\le h\le H\). 2. Differentiate: \(M'(h)=2\pi R\left(1-\frac{2h}{H}\right)\). The critical point is \(h=\frac{H}{2}\). Since \(M''(h)=-\frac{4\pi R}{H}<0\), this point gives the absolute maximum. 3. For \(R=5\) and \(H=12\), the cone’s slant height is \(s=\sqrt{5^2+12^2}=13\,\text{cm}\), so its lateral area is \(M_{\text{cone}}=\pi Rs=65\pi\,\text{cm}^2\). At \(h=6\,\text{cm}\), the cylinder radius is \(r=2.5\,\text{cm}\), and its lateral area is \(M_{\text{cyl}}=2\pi(2.5)(6)=30\pi\,\text{cm}^2\). Thus the fraction is \(\frac{M_{\text{cyl}}}{M_{\text{cone}}}=\frac{6}{13}\approx 46.2\%\).

1. \(M(h)=2\pi R\left(h-\frac{h^2}{H}\right)\). 2. The maximum occurs at \(h=\frac{H}{2}\). 3. The fraction is \(\frac{6}{13}\), or about \(46.2\%\).

- Use similar triangles to relate the rectangle’s width and height. - Express the area as a function of one dimension. - Maximize the resulting concave-down quadratic. - Substitute back to find the other dimension.

1. Similar triangles show that the horizontal width available at height \(b\) is \(a=12\left(1-\frac{b}{9}\right)\). Equivalently, \(b=9-\frac{3}{4}a\). 2. The rectangle area is \(A(a)=a\left(9-\frac{3}{4}a\right)=9a-\frac{3}{4}a^2\). 3. Differentiate: \(A'(a)=9-\frac{3}{2}a\). The critical point is \(a=6\,\text{cm}\). Since \(A''(a)=-\frac{3}{2}<0\), this point gives the maximum. 4. The corresponding height is \(b=9-\frac{3}{4}(6)=4.5\,\text{cm}\).

The area is maximized when \(a=6\,\text{cm}\) and \(b=4.5\,\text{cm}\).

- Use the cone-volume formula. - Relate the radius, height, and fixed slant height with the Pythagorean theorem. - Eliminate one variable from the volume formula. - Optimize over the physically meaningful interval.

1. The cone volume is \(V=\frac{1}{3}\pi r^2h\). Since the slant height is \(12\,\text{cm}\), \(r^2+h^2=144\), so \(r^2=144-h^2\). 2. Substitute into the volume formula: \(V(h)=\frac{\pi}{3}(144h-h^3)\), for \(0\le h\le 12\). 3. Differentiate: \(V'(h)=\frac{\pi}{3}(144-3h^2)\). The positive critical point satisfies \(h^2=48\), so \(h=4\sqrt{3}\approx 6.93\,\text{cm}\). Since \(V''(h)=-2\pi h<0\) for \(h>0\), this point gives the maximum. 4. The corresponding radius is \(r=\sqrt{144-48}=4\sqrt{6}\approx 9.80\,\text{cm}\).

The maximum-volume cone has \(r=4\sqrt{6}\approx 9.80\,\text{cm}\) and \(h=4\sqrt{3}\approx 6.93\,\text{cm}\).

- Express the perimeter using the rectangle's two side lengths. - Write the perimeter as a function of \(x\) only. - Use the derivative to find the minimizing value of \(x\). - Examine what happens to the area as \(x\) becomes very large.

1. The side lengths are \(x\) and \(f(x)\), so the perimeter is \(p(x)=2(x+f(x))=4x+\frac{16}{x}\). 2. Differentiate: \(p'(x)=4-\frac{16}{x^2}\). Setting \(p'(x)=0\) gives \(x=2\), since \(x>0\). 3. Because \(p''(x)=\frac{32}{x^3}>0\), this critical point is a minimum. Also, \(p(x)\to\infty\) as \(x\to 0^+\) and as \(x\to\infty\), so it is the global minimum. 4. The side lengths are \(x=2\) and \(f(2)=2+\frac{8}{2}=6\). 5. The minimum perimeter is \(p(2)=16\). 6. The area is \(A(x)=x f(x)=x^2+8\). Since \(A(x)\to\infty\) as \(x\to\infty\), the area has no maximum for \(x>0\).

a) The side lengths are \(2\) and \(6\). b) The minimum perimeter is \(16\). c) \(A(x)=x^2+8\), which is unbounded above, so no maximum area exists.

- Set the equations of the line and the curve equal to find the intersections. - Recall the distance formula between two points. - To minimize a square root, you can minimize its radicand. - Use the condition \(m>0\) when solving for the slope.

1. At an intersection, \(mx=\frac{9}{x}\), so \(x^2=\frac{9}{m}\). Therefore, \(x=\pm\frac{3}{\sqrt{m}}\). 2. Since \(y=mx\), the points are \(P_1=\left(\frac{3}{\sqrt{m}},3\sqrt{m}\right)\) and \(P_2=\left(-\frac{3}{\sqrt{m}},-3\sqrt{m}\right)\). 3. To minimize \(d(m)\), minimize \(h(m)=m+\frac{1}{m}\) for \(m>0\). 4. Differentiate: \(h'(m)=1-\frac{1}{m^2}\). The only critical point in the domain is \(m=1\). 5. Since \(h''(m)=\frac{2}{m^3}>0\), and \(h(m)\to\infty\) as \(m\to 0^+\) or \(m\to\infty\), this gives the global minimum. 6. When \(m=1\), the points are \(P_1=(3,3)\) and \(P_2=(-3,-3)\), and \(d(1)=\sqrt{72}=6\sqrt{2}\approx 8.49\).

a) \(P_1=\left(\frac{3}{\sqrt{m}},3\sqrt{m}\right)\) and \(P_2=\left(-\frac{3}{\sqrt{m}},-3\sqrt{m}\right)\) b) \(m=1\) c) \(P_1=(3,3)\), \(P_2=(-3,-3)\), and the minimum distance is \(6\sqrt{2}\approx 8.49\).

- Total revenue equals quantity times price. - Use the derivative to locate the highest point of the revenue function. - Verify that the critical point gives a maximum. - Report the quantity, total revenue, and price.

1. Revenue is quantity times price: \(R(x)=x(360-1.5x)=360x-1.5x^2\). 2. Differentiate: \(R'(x)=360-3x\). Setting \(R'(x)=0\) gives \(x=120\). 3. Since \(R''(x)=-3<0\), revenue is maximized at \(x=120\). 4. The maximum revenue is \(R(120)=360(120)-1.5(120)^2=21{,}600\), or \(\$21{,}600\). 5. The corresponding price is \(p(120)=360-1.5(120)=180\), or \(\$180\) per backpack.

1. \(R(x)=360x-1.5x^2\) 2. \(120\) backpacks 3. The maximum revenue is \(\$21{,}600\), at a price of \(\$180\) per backpack.

- Average cost is total cost divided by quantity. - Set the derivative of the average-cost function equal to zero. - Multiply by \(x^2\) to eliminate fractions. - Use the second derivative and end behavior to verify the minimum.

1. The average-cost function is \(C_{\text{avg}}(x)=\frac{C(x)}{x}=x^2-12x+60+\frac{256}{x}\). 2. Differentiate: \(C_{\text{avg}}'(x)=2x-12-\frac{256}{x^2}\). 3. Setting \(C_{\text{avg}}'(x)=0\) and multiplying by \(x^2\) gives \(x^3-6x^2-128=0\). 4. Factor: \(x^3-6x^2-128=(x-8)(x^2+2x+16)\). The quadratic factor has no real zeros, so the only critical value is \(x=8\). 5. Since \(C_{\text{avg}}''(x)=2+\frac{512}{x^3}>0\) for \(x>0\), and \(C_{\text{avg}}(x)\to\infty\) as \(x\to 0^+\) or \(x\to\infty\), this is the global minimum. 6. The minimum average cost is \(C_{\text{avg}}(8)=60\), or \(\$60\) per unit.

The average cost is minimized at \(x=8\) units, and the minimum average cost is \(\$60\) per unit.

- Profit equals revenue minus cost. - The company earns a profit where \(P(x)>0\). - Use the given zero to factor the cubic profit function. - Use derivatives to locate and classify the maximum.

1. Revenue is \(R(x)=52x\), so \(P(x)=R(x)-C(x)=-x^3+15x^2-48x-10\). 2. Using the given zero, factor: \(P(x)=-(x-5)(x^2-10x-2)\). 3. The other zeros are \(x=5\pm 3\sqrt{3}\). The nonnegative break-even points are \(x=5\) and \(x=5+3\sqrt{3}\approx 10.20\). 4. Profit is positive between these zeros, so the continuous-model profit interval is \((5,5+3\sqrt{3})\), approximately \((5,10.20)\). 5. Differentiate: \(P'(x)=-3x^2+30x-48=-3(x-2)(x-8)\). The derivative changes from positive to negative at \(x=8\), so profit is maximized there. 6. The maximum profit is \(P(8)=54\), or \(\$54\).

a) \(P(x)=-x^3+15x^2-48x-10\). Profit is positive for \(5<x<5+3\sqrt{3}\), approximately \(5<x<10.20\). b) Profit is maximized at \(x=8\) lamps, with a maximum profit of \(\$54\).

- Use the square-root domain condition and the first-quadrant restriction. - Express the area as width times height. - Use calculus to find the largest value of the area function. - Apply the product and chain rules when differentiating.

1. The radicand must be nonnegative: \(27-3x\ge 0\), so \(x\le 9\). In the first quadrant, \(x\ge 0\), so the interval is \([0,9]\). 2. The rectangle has width \(x\) and height \(f(x)\), so \(A(x)=x\sqrt{27-3x}\). 3. Differentiate: \(A'(x)=\frac{54-9x}{2\sqrt{27-3x}}\). 4. The derivative is zero at \(x=6\). It is positive before \(6\) and negative after \(6\), so this gives the maximum. 5. The corresponding height is \(f(6)=3\), so \(P=(6,3)\). 6. The maximum area is \(A(6)=6\cdot 3=18\).

a) \([0,9]\) b) \(P=(6,3)\), and the maximum area is \(18\) square units.

- The points have the same \(x\)-coordinate, so their distance is vertical. - Subtract the two function values to create a distance function. - Find where the derivative of the distance function is zero. - Compare all critical points and interval endpoints.

1. On \([-4,-1]\), \(f(a)\ge g(a)\), so the vertical distance is \(d(a)=f(a)-g(a)=-\frac{1}{3}a^3-2a^2-3a+3\). 2. Differentiate: \(d'(a)=-a^2-4a-3=-(a+1)(a+3)\). 3. The critical values in the interval are \(a=-3\) and \(a=-1\). 4. Evaluate the distance at all candidates: \(d(-4)=\frac{13}{3}\), \(d(-3)=3\), and \(d(-1)=\frac{13}{3}\). 5. Therefore, the minimum occurs at \(a=-3\), and the minimum distance is \(3\).

The segment is shortest when \(a=-3\), and its minimum length is \(3\).

- Use symmetry to express the full width in terms of the positive x-coordinate. - The rectangle height is the function value at its upper corner. - Write and maximize the area as a function of \(x\). - Substitute the maximizing \(x\)-value to find all requested quantities.

1. If the upper-right corner is \((x, f(x))\), symmetry gives rectangle width \(2x\) and height \(f(x)\). Thus \(A(x)=2x(-0.5x^2+6)=-x^3+12x\). 2. Differentiate: \(A'(x)=-3x^2+12\). The positive critical point satisfies \(x^2=4\), so \(x=2\). 3. Since \(A''(x)=-6x<0\) at \(x=2\), the point gives a maximum. 4. The width is \(2x=4\,\text{m}\), the height is \(f(2)=4\,\text{m}\), and the maximum area is \(4\cdot 4=16\,\text{m}^2\).

The maximum-area rectangle is \(4\,\text{m}\) wide and \(4\,\text{m}\) high, with area \(16\,\text{m}^2\).

- Marginal cost is the derivative of total cost. - Minimize a function by setting its derivative equal to zero and checking the result. - Average cost is total cost divided by quantity. - The equation in part b) can be solved by testing likely positive values.

1. The marginal-cost function is \(C'(x)=0.003x^2-0.12x+5\). 2. Differentiate marginal cost: \(C''(x)=0.006x-0.12\). Setting this equal to zero gives \(x=20\). Since \(C'''(x)=0.006>0\), marginal cost is minimized there. 3. The average-cost function is \(C_{\text{avg}}(x)=0.001x^2-0.06x+5+\frac{100}{x}\). 4. Differentiate: \(C_{\text{avg}}'(x)=0.002x-0.06-\frac{100}{x^2}\). 5. Setting \(C_{\text{avg}}'(x)=0\) gives \(0.002x^3-0.06x^2-100=0\), whose positive solution is \(x=50\). 6. Since \(C_{\text{avg}}''(x)=0.002+\frac{200}{x^3}>0\), and \(C_{\text{avg}}(x)\to\infty\) at both ends of its domain, this is the global minimum. 7. At \(x=50\), \(C'(50)=0.003(50)^2-0.12(50)+5=6.5\), and \(C_{\text{avg}}(50)=0.001(50)^2-0.06(50)+5+\frac{100}{50}=6.5\). Thus marginal cost equals average cost.

a) Marginal cost is minimized at \(x=20\) units. b) Average cost is minimized at \(x=50\) units. c) \(C'(50)=C_{\text{avg}}(50)=\$6.50\) per unit.

- Use \(V=x^2h\) to express the height in terms of \(x\). - Separate the cost of the base and lid from the cost of the side walls. - Substitute the volume constraint before differentiating. - Verify the critical point using the second derivative and end behavior.

1. The volume constraint is \(x^2h=16\), so \(h=\frac{16}{x^2}\). 2. The base and lid have total area \(2x^2\) and cost \(3(2x^2)=6x^2\). The four side walls have total area \(4xh\) and cost \(2(4xh)=8xh\). 3. Substitute \(h=\frac{16}{x^2}\): \(C(x)=6x^2+8x\left(\frac{16}{x^2}\right)=6x^2+\frac{128}{x}\). 4. Differentiate: \(C'(x)=12x-\frac{128}{x^2}\). Setting \(C'(x)=0\) gives \(12x^3=128\), so \(x=\sqrt[3]{\frac{32}{3}}\approx 2.20\,\text{ft}\). 5. Since \(C''(x)=12+\frac{256}{x^3}>0\), and \(C(x)\to\infty\) as \(x\to 0^+\) or \(x\to\infty\), this is the global minimum. 6. The corresponding height is \(h=\frac{16}{x^2}=\frac{16}{(32/3)^{2/3}}\approx 3.30\,\text{ft}\).

1. \(C(x)=6x^2+\frac{128}{x}\) 2. \(x=\sqrt[3]{\frac{32}{3}}\approx 2.20\,\text{ft}\) 3. \(h=\frac{16}{(32/3)^{2/3}}\approx 3.30\,\text{ft}\)

- Use the cylinder-volume formula. - Express the circumference in terms of the radius. - Use the mailing constraint to eliminate the tube length. - Maximize the resulting volume function and apply the requested rounding.

1. The volume is \(V=\pi r^2l\). The mailing constraint is \(l+2\pi r=120\), so \(l=120-2\pi r\). 2. Therefore, \(V(r)=\pi r^2(120-2\pi r)=120\pi r^2-2\pi^2r^3\). 3. Differentiate: \(V'(r)=240\pi r-6\pi^2r^2=6\pi r(40-\pi r)\). The positive critical point is \(r=\frac{40}{\pi}\approx 12.73\,\text{cm}\). Since \(V''\left(\frac{40}{\pi}\right)=-240\pi<0\), this point gives the maximum. 4. The corresponding length is \(l=120-2\pi\left(\frac{40}{\pi}\right)=40\,\text{cm}\). 5. The maximum volume is \(V_{\max}=\pi\left(\frac{40}{\pi}\right)^2(40)=\frac{64000}{\pi}\approx 20{,}371.8\,\text{cm}^3\).

a) \(V(r)=120\pi r^2-2\pi^2r^3\). b) \(r=\frac{40}{\pi}\approx 12.73\,\text{cm}\) and \(l=40\,\text{cm}\). c) \(V_{\max}\approx 20{,}371.8\,\text{cm}^3\).

- Use the area formula for a right triangle. - The height is the function value \(f(x)\). - Apply both the product rule and the chain rule. - Compare the critical value with the interval endpoints.

1. The triangle has base \(x\) and height \(f(x)\), so \(A(x)=\frac{1}{2}x(4-x)e^{0.5x}=(2x-0.5x^2)e^{0.5x}\). 2. Differentiate using the product and chain rules: \(A'(x)=(2-0.25x^2)e^{0.5x}\). 3. The exponential factor is positive, so \(A'(x)=0\) when \(2-0.25x^2=0\). Thus \(x^2=8\). 4. In \([0,4]\), the relevant critical value is \(x=\sqrt{8}=2\sqrt{2}\). 5. The derivative is positive before \(2\sqrt{2}\) and negative after it. Also, \(A(0)=A(4)=0\), so the absolute maximum occurs at \(x=2\sqrt{2}\approx 2.83\).

\(x=2\sqrt{2}\approx 2.83\)

- Add the rectangle area and semicircle area. - Include only exterior segments and the semicircular arc in the perimeter. - Use the perimeter constraint to eliminate \(h\). - Maximize the resulting quadratic function of \(r\).

1. The rectangle has width \(2r\). The total area is \(A=2rh+\frac{1}{2}\pi r^2\). The outside perimeter is \(2h+2r+\pi r=6\), so \(h=3-r\left(1+\frac{\pi}{2}\right)\). 2. Substitute into the area: \(A(r)=6r-r^2\left(2+\frac{\pi}{2}\right)\). 3. Differentiate: \(A'(r)=6-(4+\pi)r\). The critical point is \(r=\frac{6}{4+\pi}\approx 0.840\,\text{m}\). Since \(A''(r)=-(4+\pi)<0\), this point gives the maximum. 4. The rectangular height is \(h=3-\frac{6}{4+\pi}\left(1+\frac{\pi}{2}\right)=\frac{6}{4+\pi}\approx 0.840\,\text{m}\).

The area is maximized when \(r=h=\frac{6}{4+\pi}\approx 0.840\,\text{m}\).

- Express the budget in units of side-wall material cost. - Weight the square base by a factor of \(3\). - Use the budget constraint to eliminate the height. - Maximize the resulting one-variable volume function.

1. Measured in equivalent square meters of side-wall material, the budget constraint is \(3a^2+4ah=9\). Thus \(h=\frac{9-3a^2}{4a}\). 2. The volume is \(V(a)=a^2h=\frac{9a-3a^3}{4}\), for \(0<a<\sqrt{3}\). 3. Differentiate: \(V'(a)=\frac{9}{4}-\frac{9}{4}a^2\). The positive critical point is \(a=1\,\text{m}\). Since \(V''(a)=-\frac{9}{2}a<0\) for \(a>0\), this point gives the maximum. 4. The height is \(h=\frac{9-3(1)^2}{4(1)}=1.5\,\text{m}\), and the maximum volume is \(V=1^2(1.5)=1.5\,\text{m}^3\).

a) \(a=1\,\text{m}\) and \(h=1.5\,\text{m}\). b) \(V_{\max}=1.5\,\text{m}^3\).

- Let one sheet dimension represent the cylinder circumference and the other the height. - Use the sheet perimeter to eliminate one dimension. - Convert the circumference to a radius before writing the volume. - Maximize the resulting one-variable function.

1. Let \(x\) be the side that becomes the base circumference and let \(h\) be the cylinder height. The sheet-perimeter constraint is \(2x+2h=120\), so \(h=60-x\). Since \(x=2\pi r\), \(r=\frac{x}{2\pi}\). 2. The cylinder volume is \(V(x)=\pi\left(\frac{x}{2\pi}\right)^2(60-x)=\frac{60x^2-x^3}{4\pi}\), for \(0\le x\le 60\). 3. Differentiate: \(V'(x)=\frac{120x-3x^2}{4\pi}=\frac{3x(40-x)}{4\pi}\). The interior critical point is \(x=40\,\text{cm}\). Since \(V''(40)=\frac{-120}{4\pi}<0\), this point gives the maximum. 4. The other sheet dimension is \(h=60-40=20\,\text{cm}\).

The sheet should measure \(40\,\text{cm}\times 20\,\text{cm}\), with the \(40\,\text{cm}\) side forming the circular circumference.

- Use the net dimensions to express the base length and width in terms of \(h\). - Determine the physical domain for the box height. - Write and differentiate the volume function. - Discard any critical point outside the domain and verify the maximum.

1. Let \(l\) and \(w\) be the base dimensions. The net constraints are \(2l+2h=80\Rightarrow l=40-h\) and \(w+2h=50\Rightarrow w=50-2h\). 2. The volume is \(V(h)=h(40-h)(50-2h)=2h^3-130h^2+2000h\). Positive dimensions require \(0<h<25\). 3. Differentiate: \(V'(h)=6h^2-260h+2000\). Solving \(V'(h)=0\) gives \(h=10\) or \(h=\frac{100}{3}\). Only \(h=10\) lies in the physical domain. 4. Since \(V''(10)=12(10)-260=-140<0\), the critical point gives a maximum. The volume approaches \(0\) at both ends of the physical domain, so it is the absolute maximum. 5. The maximum volume is \(V(10)=10(30)(30)=9000\,\text{cm}^3\).

The box height should be \(h=10\,\text{cm}\), and the maximum volume is \(9000\,\text{cm}^3\).

- Add the flat bottom, cylindrical side, and curved hemispherical area. - Add the cylinder and hemisphere volumes. - Use the volume constraint to eliminate \(h\). - Minimize the resulting surface-area function.

1. The exterior area is \(S=\pi r^2+2\pi rh+2\pi r^2=3\pi r^2+2\pi rh\). The volume constraint is \(\pi r^2h+\frac{2}{3}\pi r^3=1.5\). Thus \(h=\frac{1.5}{\pi r^2}-\frac{2}{3}r\). 2. Substitute into the area: \(S(r)=\frac{5}{3}\pi r^2+\frac{3}{r}\), for \(r>0\). 3. Differentiate: \(S'(r)=\frac{10}{3}\pi r-\frac{3}{r^2}\). The critical point satisfies \(r^3=\frac{9}{10\pi}\), so \(r=\sqrt[3]{\frac{9}{10\pi}}\approx 0.659\,\text{m}\). 4. Since \(S''(r)=\frac{10}{3}\pi+\frac{6}{r^3}>0\), the critical point gives a minimum. Using the volume constraint and the critical-point equation gives \(h=r\approx 0.659\,\text{m}\).

The minimum-material dimensions are \(r=\sqrt[3]{\frac{9}{10\pi}}\approx 0.659\,\text{m}\) and \(h\approx 0.659\,\text{m}\).

- Write the fixed-cost constraint using the two material rates. - Solve the cost constraint for the height. - Substitute into the volume and find the critical point. - Use the critical-point equation in the original cost constraint to determine the ratio.

1. Let \(k\) be the side-material cost per unit area and \(C\) be the fixed total cost. The cost constraint is \(C=2k\pi r^2+2k\pi rh\). Solving for \(h\) gives \(h=\frac{C}{2k\pi r}-r\). 2. The volume is \(V(r)=\pi r^2h=\frac{C}{2k}r-\pi r^3\). 3. Differentiate: \(V'(r)=\frac{C}{2k}-3\pi r^2\). At the critical point, \(C=6k\pi r^2\). Substitute this into the original cost constraint: \(6k\pi r^2=2k\pi r^2+2k\pi rh\). Therefore, \(h=2r\). The cubic volume function is concave down at this critical point, so it gives the maximum.

The optimal ratio is \(h:r=2:1\).

- Identify the quantity to maximize and write its formula. - Use the fixed surface area to express one variable in terms of the other. - Find the critical point of the one-variable volume function. - Check that the dimensions are physically meaningful.

1. The volume is \(V=\pi r^2h\). The fixed surface-area condition is \(2\pi r^2+2\pi rh=600\). 2. Solve the constraint for \(h\): \(h=\frac{300}{\pi r}-r\). 3. Substitute into the volume: \(V(r)=\pi r^2\left(\frac{300}{\pi r}-r\right)=300r-\pi r^3\). 4. Differentiate and set the derivative equal to zero: \(V'(r)=300-3\pi r^2=0\). Thus \(r=\sqrt{\frac{100}{\pi}}\approx 5.64\,\text{cm}\). 5. Since \(V''(r)=-6\pi r<0\) for \(r>0\), the critical point is a maximum. Also, the volume approaches \(0\) at both ends of the physical domain, so this is the absolute maximum. 6. The height is \(h=\frac{300}{\pi r}-r=2r\approx 11.28\,\text{cm}\). 7. The maximum volume is \(V=\pi r^2h\approx 1128.38\,\text{cm}^3\).

The can should have radius \(r\approx 5.64\,\text{cm}\) and height \(h\approx 11.28\,\text{cm}\). Its maximum volume is \(V\approx 1128.38\,\text{cm}^3\).

- Write formulas for the fixed volume and the weighted material cost. - Use the volume condition to eliminate either \(h\) or \(r\). - Express the cost as a function of one variable. - Use derivatives to find the minimum.

1. The fixed volume is \(V=\pi r^2h+\frac{2}{3}\pi r^3\). 2. Let the side material cost one unit per square meter. Then the weighted cost is \(C(r, h)=2\pi rh+2(2\pi r^2)=2\pi rh+4\pi r^2\). 3. Solve the volume equation for \(h\): \(h=\frac{V}{\pi r^2}-\frac{2}{3}r\). 4. Substitute into the cost function: \(C(r)=2\pi r\left(\frac{V}{\pi r^2}-\frac{2}{3}r\right)+4\pi r^2=\frac{2V}{r}+\frac{8}{3}\pi r^2\). 5. Differentiate and set the derivative equal to zero: \(C'(r)=-\frac{2V}{r^2}+\frac{16}{3}\pi r=0\). Thus \(V=\frac{8}{3}\pi r^3\). 6. Substitute this relation into the expression for \(h\): \(h=\frac{\frac{8}{3}\pi r^3}{\pi r^2}-\frac{2}{3}r=2r\). 7. Since \(C''(r)=\frac{4V}{r^3}+\frac{16}{3}\pi>0\), the cost is minimized at this critical point.

The optimal ratio is \(h:r=2:1\). The cylindrical height should equal the diameter of the hemispherical bottom.

- Sketch the symmetric rectangle under the parabola. - Express both width and height using the positive \(x\)-coordinate of an upper vertex. - Write the area as a function of \(x\). - Determine the allowable interval for \(x\). - Use the derivative to maximize the area.

1. By symmetry, let the upper-right vertex be \((x,12-x^2)\), where \(0<x<\sqrt{12}\). The rectangle has width \(2x\) and height \(12-x^2\). 2. Its area is \(A(x)=2x(12-x^2)=24x-2x^3\). 3. Differentiate: \(A'(x)=24-6x^2\). Setting the derivative equal to zero gives \(x=2\). 4. Since \(A''(2)=-24<0\), this critical point is a local maximum. The area approaches \(0\) at both ends of the domain, so it is the global maximum. 5. The width is \(2x=4\), and the height is \(12-x^2=8\).

The side lengths are \(4\) and \(8\).

- Use a vertical cross section through the center of the pyramid. - Express the prism volume as a function of one variable. - Find the interior critical point of that volume function. - Compare the maximum prism volume with \(\frac{1}{3}a^2h\).

1. Take a vertical cross section through the pyramid's apex and the midpoints of two opposite base edges. The large triangle has base \(a\) and height \(h\). The smaller triangle above the prism has base \(x\) and height \(h-y\). By similarity, \(\frac{h-y}{h}=\frac{x}{a}\), so \(y=h\left(1-\frac{x}{a}\right)\). 2. The prism volume is \(V(x)=x^2y=hx^2-\frac{h}{a}x^3\). Differentiate: \(V'(x)=2hx-\frac{3h}{a}x^2\). For \(0<x<a\), the critical point satisfies \(2h-\frac{3h}{a}x=0\), giving \(x=\frac{2}{3}a\). Then \(y=h\left(1-\frac{2}{3}\right)=\frac{1}{3}h\). Since \(V''\left(\frac{2a}{3}\right)=-2h<0\), this critical point gives the maximum. 3. The maximum prism volume is \(V_{\max}=\left(\frac{2a}{3}\right)^2\left(\frac{h}{3}\right)=\frac{4}{27}a^2h\). The pyramid volume is \(V_{\text{pyramid}}=\frac{1}{3}a^2h\). Therefore, \(\frac{V_{\max}}{V_{\text{pyramid}}}=\frac{4}{9}\).

1. Similar triangles give \(y=h\left(1-\frac{x}{a}\right)\). 2. The maximizing dimensions are \(x=\frac{2}{3}a\) and \(y=\frac{1}{3}h\). 3. The prism occupies \(\frac{4}{9}\) of the pyramid's volume, or about \(44.4\%\).

- Draw a cross section through a base diagonal and the apex. - Use the fixed lateral edge to express the base area in terms of \(h\). - Determine the possible interval for the height. - Differentiate the resulting volume function.

1. Let \(a\) be the base edge length. The volume is \(V=\frac{1}{3}a^2h\). 2. A right triangle formed by the height, half the base diagonal, and a lateral edge gives \(9^2=h^2+\left(\frac{a\sqrt{2}}{2}\right)^2\). Therefore, \(a^2=2(81-h^2)\). 3. Substitute into the volume: \(V(h)=\frac{2}{3}(81h-h^3)\), where \(0<h<9\). 4. Differentiate: \(V'(h)=\frac{2}{3}(81-3h^2)\). The positive critical point satisfies \(h^2=27\), so \(h=3\sqrt{3}\,\text{cm}\approx 5.20\,\text{cm}\). 5. Since \(V''(h)=-4h<0\) for \(h>0\), and the volume is \(0\) at the endpoints of the physical domain, this point gives the absolute maximum. 6. The maximum volume is \(V(3\sqrt{3})=108\sqrt{3}\,\text{cm}^3\approx 187.06\,\text{cm}^3\).

The maximizing height is \(h=3\sqrt{3}\,\text{cm}\approx 5.20\,\text{cm}\). The maximum volume is \(108\sqrt{3}\,\text{cm}^3\approx 187.06\,\text{cm}^3\).

- Write the cylinder volume formula. - Use an axial cross section to relate \(r\), \(h\), and the sphere radius. - Eliminate one variable from the volume formula. - Include the endpoints of the physical domain in your maximum check.

1. The cylinder volume is \(V=\pi r^2h\). In an axial cross section of the sphere, \(r^2+\left(\frac{h}{2}\right)^2=6^2\). Thus \(r^2=36-\frac{h^2}{4}\). 2. Substitute into the volume: \(V(h)=\pi\left(36-\frac{h^2}{4}\right)h=36\pi h-\frac{\pi}{4}h^3\), where \(0\le h\le12\). 3. Differentiate: \(V'(h)=36\pi-\frac{3\pi}{4}h^2\). Setting the derivative equal to zero gives \(h^2=48\), so \(h=4\sqrt{3}\,\text{m}\approx 6.93\,\text{m}\). 4. Since \(V''(h)=-\frac{3\pi}{2}h<0\) at the positive critical point, and the endpoint volumes are \(0\), this is the absolute maximum. 5. The radius is \(r=\sqrt{36-\frac{48}{4}}=2\sqrt{6}\,\text{m}\approx 4.90\,\text{m}\). 6. The maximum volume is \(V=\pi(24)(4\sqrt{3})=96\sqrt{3}\pi\,\text{m}^3\approx 522.37\,\text{m}^3\).

The cylinder has radius \(r\approx 4.90\,\text{m}\) and height \(h\approx 6.93\,\text{m}\). Its maximum volume is \(V\approx 522.37\,\text{m}^3\).

- Write the volume in terms of \(a\) and \(h\). - Use the fixed space diagonal to eliminate one variable. - Differentiate the one-variable volume function. - Compare the resulting values of \(a\) and \(h\).

1. The volume is \(V=a^2h\). The space-diagonal constraint is \(D^2=2a^2+h^2\). Therefore, \(a^2=\frac{D^2-h^2}{2}\). 2. Substitute into the volume: \(V(h)=\frac{1}{2}(D^2h-h^3)\), where \(0<h<D\). 3. Differentiate: \(V'(h)=\frac{1}{2}(D^2-3h^2)\). The positive critical point is \(h=\frac{D}{\sqrt{3}}\). 4. Then \(a^2=\frac{D^2-D^2/3}{2}=\frac{D^2}{3}\), so \(a=\frac{D}{\sqrt{3}}\). The volume is zero at both endpoints of the physical domain, so the interior critical point gives the maximum. 5. Therefore, \(\frac{h}{a}=1\).

The optimal ratio is \(\frac{h}{a}=1\). The maximum-volume box is a cube.

- Revenue equals quantity times price. - Profit equals revenue minus cost. - At an interior profit maximum, the derivative is zero. - Use the second derivative to verify the maximum.

1. Revenue is \(R(x)=x p(x)=-0.4x^2+84x\). 2. Profit is \(P(x)=R(x)-C(x)=-0.4x^2+72x-400\). 3. Differentiate: \(P'(x)=-0.8x+72\). Setting the derivative equal to zero gives \(x=90\). 4. Since \(P''(x)=-0.8<0\), profit is maximized at \(x=90\). 5. The corresponding price is \(p(90)=-0.4(90)+84=48\), or \(\$48\). 6. The maximum profit is \(P(90)=-0.4(90)^2+72(90)-400=2840\), or \(\$2840\).

Profit is maximized at \(x=90\) units. The corresponding price is \(\$48\) per unit, and the maximum profit is \(\$2840\).

- Revenue equals quantity times price. - Use the product and chain rules to differentiate the exponential revenue function. - The exponential factor never equals zero. - Include units and round monetary values to two decimal places.

1. Revenue is \(R(x)=x p(x)=300x e^{-0.005x}\). 2. Differentiate using the product and chain rules: \(R'(x)=300e^{-0.005x}(1-0.005x)\). 3. The exponential factor is positive, so \(R'(x)=0\) when \(1-0.005x=0\). Thus \(x=200\). 4. The derivative changes from positive to negative at \(x=200\), so revenue is maximized there. 5. The corresponding price is \(p(200)=300e^{-1}\approx \$110.36\). 6. The maximum revenue is \(R(200)=60{,}000e^{-1}\approx \$22{,}072.77\).

Revenue is maximized at \(x=200\) units. The price is approximately \(\$110.36\) per unit, and the maximum revenue is approximately \(\$22{,}072.77\).

- The selling price depends on the number of bags sold. - Revenue equals quantity times the price-demand function. - Write the complete profit function before differentiating. - Evaluate profit at the maximizing production level.

1. Revenue is \(R(x)=x p(x)=300x-10x^2\). 2. Profit is \(P(x)=R(x)-C(x)=-x^3+5x^2+200x-500\). 3. Differentiate: \(P'(x)=-3x^2+10x+200\). Setting the derivative equal to zero gives \(x=10\) or \(x=-\frac{20}{3}\). 4. Only \(x=10\) is in the production domain. Since \(P''(10)=-50<0\), profit is maximized there. 5. The maximum profit is \(P(10)=-1000+500+2000-500=1000\), or \(\$1000\).

Profit is maximized when \(10\) bags are produced and sold, and the maximum profit is \(\$1000\).

- Represent a general point on the parabola as \((x, x^2)\). - Minimize the squared distance to avoid a square root. - Try small integer values when solving the cubic equation. - Use the second derivative to verify the minimum.

1. A point on the graph is \((x, x^2)\). Minimize the squared distance \(D(x)=(x-3)^2+x^4=x^4+x^2-6x+9\). 2. Differentiate: \(D'(x)=4x^3+2x-6\). The equation \(D'(x)=0\) has the solution \(x=1\). 3. Since \(D''(x)=12x^2+2>0\) for every real \(x\), \(D\) is strictly convex. Therefore, its only critical point is the absolute minimum. 4. The corresponding y-coordinate is \(f(1)=1\). Thus the closest point is \((1, 1)\).

The closest point is \(P(1, 1)\).

- Use the given relationship between length and width. - Write the total surface area of a closed rectangular box. - Use the fixed volume to express \(h\) in terms of \(b\). - Find the minimum of the resulting surface-area function.

1. Let \(l=3b\). The surface area is \(O=2(lb+lh+bh)=6b^2+8bh\). The volume constraint is \(lbh=3b^2h=36\). 2. Solve for \(h\): \(h=\frac{12}{b^2}\). Substitute into the surface area: \(O(b)=6b^2+\frac{96}{b}\). 3. Differentiate: \(O'(b)=12b-\frac{96}{b^2}\). Setting the derivative equal to zero gives \(12b^3=96\), so \(b=2\,\text{ft}\). 4. Then \(l=3b=6\,\text{ft}\) and \(h=\frac{12}{2^2}=3\,\text{ft}\). 5. Since \(O''(b)=12+\frac{192}{b^3}>0\) for \(b>0\), and \(O(b)\to\infty\) at both ends of the positive domain, these dimensions give the absolute minimum.

The container dimensions are \(l=6\,\text{ft}\), \(b=2\,\text{ft}\), and \(h=3\,\text{ft}\).

- Substitute the fixed values before optimizing. - Rewrite \(\frac{1}{\cos^2(\alpha)}\) using \(\tan(\alpha)\). - Let \(t=\tan(\alpha)\) to obtain a quadratic objective function. - Use the vertex or derivative of that quadratic.

1. For \(\alpha=45^\circ\), \(y(30)=30\tan(45^\circ)-\frac{10(30)^2}{2(20)^2\cos^2(45^\circ)}\). Using \(\tan(45^\circ)=1\) and \(\cos^2(45^\circ)=0.5\), \(y(30)=30-22.5=7.5\,\text{m}\). 2. At the wall, the height as a function of angle is \(h(\alpha)=30\tan(\alpha)-\frac{11.25}{\cos^2(\alpha)}\). Let \(t=\tan(\alpha)\). Since \(\sec^2(\alpha)=1+t^2\), \(h(t)=30t-11.25(1+t^2)\), where \(t\ge0\). 3. Differentiate: \(h'(t)=30-22.5t\). The critical point is \(t=\frac{4}{3}\). Because the quadratic opens downward, this point gives the absolute maximum. 4. Thus \(\alpha=\arctan\left(\frac{4}{3}\right)\approx53.13^\circ\). The maximum height is \(h_{\max}=30\left(\frac{4}{3}\right)-11.25\left(1+\frac{16}{9}\right)=8.75\,\text{m}\).

a) The capsule hits the wall at a height of \(7.5\,\text{m}\). b) The maximum impact height is \(8.75\,\text{m}\), reached at \(\alpha\approx53.13^\circ\).

- Marginal cost is the derivative of total cost. - Average cost is total cost divided by quantity. - Use derivative conditions to locate minima and maxima. - Profit equals revenue minus cost. - Test likely integer roots for the average-cost equation.

1. Marginal cost is \(C'(x)=3x^2-18x+30\). Its derivative is \(C''(x)=6x-18\), which is zero at \(x=3\). Since \(C'''(x)=6>0\), marginal cost is minimized at \(x=3\). 2. Average cost is \(C_{\text{avg}}(x)=\frac{C(x)}{x}=x^2-9x+30+\frac{25}{x}\). 3. Differentiate: \(C_{\text{avg}}'(x)=2x-9-\frac{25}{x^2}\). Setting this equal to zero gives \(2x^3-9x^2-25=0\), which has the positive solution \(x=5\). Since \(C_{\text{avg}}''(x)=2+\frac{50}{x^3}>0\) for \(x>0\), average cost is minimized at \(x=5\). 4. Revenue is \(R(x)=51x\), so profit is \(P(x)=R(x)-C(x)=-x^3+9x^2+21x-25\). 5. Differentiate: \(P'(x)=-3x^2+18x+21=-3(x-7)(x+1)\). The positive critical value is \(x=7\), and \(P''(7)=-24<0\), so profit is maximized at \(x=7\).

a) \(x=3\) units b) \(x=5\) units c) \(x=7\) units

- Separate the expression into one factor involving \(T_1\) and one involving \(T_2\). - Maximize each one-variable factor on its interval. - Why does maximizing both nonnegative factors maximize their product? - Include the interval endpoints in each comparison.

1. Write \(A(T_1, T_2) = F(T_1)G(T_2)\), where \(F(T_1) = 3T_1^2 - T_1^3\) and \(G(T_2) = 12T_2 - T_2^2\). Both factors are nonnegative on the stated intervals, so the product is maximized when both factors are maximized. 2. Differentiate \(F\): \(F'(T_1) = 6T_1 - 3T_1^2 = 3T_1(2 - T_1)\). The candidates are \(T_1 = 0\), \(T_1 = 2\), and the endpoint \(T_1 = 3\). Since \(F(0) = F(3) = 0\) and \(F(2) = 4\), the maximum occurs at \(T_1 = 2\). 3. Differentiate \(G\): \(G'(T_2) = 12 - 2T_2\). The candidates are \(T_2 = 6\) and the endpoints. Since \(G(0) = G(12) = 0\) and \(G(6) = 36\), the maximum occurs at \(T_2 = 6\). 4. Therefore, the yield is maximized when \(T_1 = 2\) and \(T_2 = 6\).

\(T_1 = 2\) and \(T_2 = 6\)

- Write the sector boundary as two radii plus one arc. - Use \(A=\frac{1}{2}rb\) for sector area. - Eliminate the arc length and maximize the resulting quadratic. - Use \(b=r\varphi\) to find the angle.

1. The boundary-length constraint is \(2r+b=20\), so \(b=20-2r\). 2. The sector area can be written as \(A=\frac{1}{2}rb\). Therefore, \(A(r)=\frac{1}{2}r(20-2r)=10r-r^2\). 3. Differentiate: \(A'(r)=10-2r\). The critical point is \(r=5\,\text{cm}\). Since \(A''(r)=-2<0\), this point gives the maximum. 4. The arc length is \(b=20-2(5)=10\,\text{cm}\). Using \(b=r\varphi\), \(10=5\varphi\), so \(\varphi=2\) radians.

The maximum-area sector has radius \(r=5\,\text{cm}\) and central angle \(\varphi=2\) radians.

- Draw the coastline, station, platform, and entry point. - Use the Pythagorean theorem for the underwater distance. - Build a total-cost function using the two rates. - Compare the critical point with both endpoints.

1. Let \(x\) be the distance in miles from \(A\) to \(P\) along the coast. The land distance is \(x\), and the underwater distance is \(\sqrt{(15-x)^2+8^2}\). 2. The cost function is \(K(x)=30{,}000x+50{,}000\sqrt{(15-x)^2+64}\), where \(0\le x\le15\). 3. Differentiate: \(K'(x)=30{,}000-\frac{50{,}000(15-x)}{\sqrt{(15-x)^2+64}}\). Setting the derivative equal to zero gives \(3\sqrt{(15-x)^2+64}=5(15-x)\). Squaring and simplifying yields \(15-x=6\), so \(x=9\). 4. Compare the critical point and endpoints: \(K(0)=\$850{,}000\), \(K(15)=\$850{,}000\), and \(K(9)=30{,}000(9)+50{,}000(10)=\$770{,}000\). Therefore, the critical point gives the minimum.

The pipeline should enter the water \(9\) miles east of station \(A\). The minimum total cost is \(\$770{,}000\).

- How is the annual increase \(G(x)\) related to the reproduction model \(f(x)\)? - What derivative condition holds at an interior maximum? - How can the population value for the following year be written as an equation involving \(f\)? - Use the two conditions to create a system for \(a\) and \(b\).

1. Write the annual-increase function: \(G(x)=f(x)-x=ax^2+(b-1)x\). 2. Since \(f(400)=500\), \(160{,}000a+400b=500\). 3. The increase is maximized at \(x=400\), so \(G'(400)=0\). Because \(G'(x)=2ax+b-1\), this gives \(800a+b=1\). 4. From the second equation, \(b=1-800a\). Substitute into the first equation: \(160{,}000a+400(1-800a)=500\). 5. Simplify: \(-160{,}000a=100\), so \(a=-\frac{1}{1600}=-0.000625\). 6. Then \(b=1-800\left(-\frac{1}{1600}\right)=\frac{3}{2}=1.5\). 7. Since \(G''(x)=2a=-\frac{1}{800}<0\), the critical point is a maximum, as required.

\(a=-\frac{1}{1600}=-0.000625\) and \(b=\frac{3}{2}=1.5\)

- A rectangular box has four edges of each dimension. - Use \(l=1.5w\) in the edge-sum constraint. - Use the height restriction to determine the domain for \(w\). - Compare the critical point with the boundary value and boundary behavior.

1. The volume is \(V=wlh=1.5w^2h\). The edge-length constraint is \(4(w+l+h)=240\). Using \(l=1.5w\), \(2.5w+h=60\), so \(h=60-2.5w\). 2. The volume function is \(V(w)=1.5w^2(60-2.5w)=90w^2-3.75w^3\). The height restriction \(h\le30\) gives \(w\ge12\), while \(h>0\) gives \(w<24\). Thus \(12\le w<24\). 3. Differentiate: \(V'(w)=180w-11.25w^2\). The interior critical point is \(w=16\). 4. Compare the critical point and boundary behavior: \(V(12)=6480\,\text{in}^3\), \(V(16)=7680\,\text{in}^3\), and \(V(w)\to0\) as \(w\to24^-\). Therefore, the absolute maximum occurs at \(w=16\,\text{in}\). 5. The remaining dimensions are \(l=1.5(16)=24\,\text{in}\) and \(h=60-2.5(16)=20\,\text{in}\).

a) The box should have width \(16\,\text{in}\), length \(24\,\text{in}\), and height \(20\,\text{in}\). b) The maximum volume is \(7680\,\text{in}^3\).

- Use symmetry to express the width as \(2x\). - The height is the function value \(f(x)\). - Differentiate the resulting rational area function. - Report the rectangle's side lengths, not just the value of \(x\).

1. Let the upper vertices be \((\pm x, f(x))\), where \(x>0\). The width is \(2x\), and the height is \(f(x)\). Thus \(A(x)=2x\left(\frac{4}{x^2+1}\right)=\frac{8x}{x^2+1}\). 2. Differentiate: \(A'(x)=\frac{8(1-x^2)}{(x^2+1)^2}\). The only positive critical point is \(x=1\). 3. The derivative is positive for \(0<x<1\) and negative for \(x>1\). Also, the area approaches \(0\) at both ends of the positive domain, so \(x=1\) gives the absolute maximum. 4. The width is \(2x=2\), and the height is \(f(1)=\frac{4}{2}=2\).

The maximum-area rectangle has side lengths \(2\) and \(2\).

- Find the secant slope from the two given points. - Subtract the function from the secant line to form the vertical-distance function. - Find the critical point of that difference. - Check the interval endpoints.

1. The endpoint values are \(f(1)=4\) and \(f(4)=1\). The secant slope is \(m=\frac{1-4}{4-1}=-1\). Thus \(s(x)=-x+5\). 2. The vertical distance is \(d(x)=s(x)-f(x)=-x+5-\frac{4}{x}\). 3. Differentiate: \(d'(x)=-1+\frac{4}{x^2}\). Setting the derivative equal to zero gives \(x=2\). 4. Since \(d''(x)=-\frac{8}{x^3}<0\) on the interval, the critical point gives a maximum. Also, \(d(1)=d(4)=0\), so it is the absolute maximum. 5. The maximum distance is \(d(2)=-2+5-2=1\).

a) \(s(x)=-x+5\) b) The maximum vertical distance occurs at \(x=2\), and the distance is \(1\).

- Represent a general point on the graph as \((x, e^x)\). - Minimize the squared distance to avoid a square root. - Test a simple value such as \(x=0\) in the critical-point equation. - Use the second derivative to establish uniqueness.

1. A general point on the graph is \((x, e^x)\). Minimize the squared distance \(g(x)=(x-1)^2+e^{2x}\). 2. Differentiate: \(g'(x)=2(x-1)+2e^{2x}\). At \(x=0\), \(g'(0)=-2+2=0\). 3. The second derivative is \(g''(x)=2+4e^{2x}>0\) for every real \(x\). Thus \(g\) is strictly convex, so this critical point is the unique absolute minimum. 4. Since \(f(0)=1\), the closest point is \((0, 1)\).

The closest point is \(Q(0, 1)\).

- Express the rectangle's area using the coordinates of its graph point. - Simplify the area function before differentiating. - Use the second derivative and end behavior to verify the absolute minimum. - Report both coordinates.

1. If the vertex is \((x, f(x))\), the area is \(A(x)=xf(x)=x^2+\frac{54}{x}\). 2. Differentiate: \(A'(x)=2x-\frac{54}{x^2}\). Setting the derivative equal to zero gives \(2x^3=54\), so \(x=3\). 3. Since \(A''(x)=2+\frac{108}{x^3}>0\) for \(x>0\), the area function is strictly convex. Also, \(A(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), so the critical point gives the absolute minimum. 4. The y-coordinate is \(f(3)=\frac{27+54}{9}=9\).

The area is minimized when the vertex is \((3, 9)\).

- Write the line through \(P\) using its slope. - Express both intercepts in terms of the slope. - Use the condition that both intercepts are positive. - Minimize their sum and check the end behavior of the domain.

1. Let the line have slope \(m<0\). Its equation through \(P\) is \(y-1=m(x-4)\). The intercepts are \(x_0=4-\frac{1}{m}\) and \(y_0=1-4m\). 2. Their sum is \(S(m)=5-\frac{1}{m}-4m\), where \(m<0\). 3. Differentiate: \(S'(m)=\frac{1}{m^2}-4\). Setting the derivative equal to zero gives \(m=-\frac{1}{2}\). 4. Since \(S''(m)=-\frac{2}{m^3}>0\) at \(m=-\frac{1}{2}\), and \(S(m)\to\infty\) as \(m\to0^-\) or \(m\to-\infty\), this point gives the absolute minimum. 5. Therefore, \(x_0=4-\frac{1}{-1/2}=6\) and \(y_0=1-4\left(-\frac{1}{2}\right)=3\).

The minimizing intercepts are \(x_0=6\) and \(y_0=3\).

- Draw the rectangle inside the isosceles triangle. - Use similar triangles to relate the triangle base and height. - Determine the domain for \(h\). - Minimize the one-variable area function.

1. Let \(g\) be the triangle's base length. Similar triangles give \(\frac{h-4}{10}=\frac{h}{g}\). Thus \(g=\frac{10h}{h-4}\), where \(h>4\). 2. The triangle area is \(A(h)=\frac{1}{2}gh=\frac{5h^2}{h-4}\). 3. Differentiate: \(A'(h)=\frac{5h(h-8)}{(h-4)^2}\). The only critical point in the domain is \(h=8\,\text{cm}\). 4. The derivative changes from negative to positive at \(h=8\). Also, \(A(h)\to\infty\) as \(h\to4^+\) or \(h\to\infty\), so this is the absolute minimum.

The triangle's area is minimized when its height is \(h=8\,\text{cm}\).

- Use the volume and surface-area formulas for a cylinder. - Use the fixed volume to express height in terms of radius. - Differentiate the one-variable cost function. - Use the second derivative to verify a minimum.

1. The volume constraint is \(\pi r^2h=27\pi\), so \(h=\frac{27}{r^2}\). 2. The base and lid cost \(2\pi r^2\) cents. The curved side has area \(2\pi rh\) and costs \(4\pi rh\) cents. Thus \(C(r,h)=2\pi r^2+4\pi rh\). 3. Substitute the volume constraint: \(C(r)=2\pi r^2+\frac{108\pi}{r}\). 4. Differentiate: \(C'(r)=4\pi r-\frac{108\pi}{r^2}\). 5. Setting \(C'(r)=0\) gives \(4\pi r^3=108\pi\), so \(r=3\,\text{in}\). 6. Then \(h=\frac{27}{3^2}=3\,\text{in}\). 7. The second derivative is \(C''(r)=4\pi+\frac{216\pi}{r^3}\), and \(C''(3)=12\pi>0\). Therefore, the dimensions give a cost minimum.

The material cost is minimized when \(r=3\,\text{in}\) and \(h=3\,\text{in}\).

- Use symmetry to find the triangle's base length. - Write the area as one-half times base times height. - Apply the product and chain rules. - Give both the maximizing x-value and the coordinates of \(P\).

1. The triangle has base \(2x\) and height \(f(x)\), so \(A(x)=\frac{1}{2}(2x)(6e^{-0.1x^2})=6xe^{-0.1x^2}\). 2. Differentiate using the product and chain rules: \(A'(x)=(6-1.2x^2)e^{-0.1x^2}\). Since the exponential factor is positive, the critical point satisfies \(6-1.2x^2=0\). Thus \(x=\sqrt{5}\approx2.24\). 3. The derivative changes from positive to negative at \(x=\sqrt{5}\). Also, the area approaches zero as \(x\to0^+\) or \(x\to\infty\), so this is the absolute maximum. 4. The y-coordinate is \(f(\sqrt{5})=6e^{-0.5}\approx3.64\).

The area is maximized at \(x=\sqrt{5}\approx2.24\). The point is \(P(\sqrt{5}, 6e^{-0.5})\approx(2.24, 3.64)\).

- Add the cylinder and hemisphere volumes. - Use the fixed volume to express \(h\) in terms of \(r\). - Weight each surface area by its material cost. - Recall the curved surface area of a hemisphere and the lateral area of a cylinder. - Differentiate the one-variable cost function.

1. The fixed volume is \(V=\pi r^2h+\frac{2}{3}\pi r^3\). 2. Solve for the cylindrical height: \(h=\frac{V}{\pi r^2}-\frac{2}{3}r\). 3. Let the unit cost for the side and lid be \(k\). The cylindrical side and lid cost \(k(2\pi rh+\pi r^2)\). The hemispherical bottom has curved area \(2\pi r^2\) and unit cost \(2k\), so its cost is \(4k\pi r^2\). 4. Ignoring the positive common factor \(k\), the cost function is \(C(r)=2\pi rh+5\pi r^2\). 5. Substitute for \(h\): \(C(r)=\frac{2V}{r}+\frac{11}{3}\pi r^2\). 6. Differentiate: \(C'(r)=-\frac{2V}{r^2}+\frac{22}{3}\pi r\). Setting this equal to zero gives \(V=\frac{11}{3}\pi r^3\). 7. Substitute into the height formula: \(h=\frac{11}{3}r-\frac{2}{3}r=3r\). 8. Since \(C''(r)=\frac{4V}{r^3}+\frac{22}{3}\pi>0\), this gives the minimum cost.

\(h:r=3:1\)

- Find the first-quadrant intercepts before integrating. - Use a definite integral for the enclosed area. - Write the rectangle area as \(u f(u)\). - Keep only critical points in \((0, 2)\).

1. The graph meets the y-axis at \((0, 16)\) and the x-axis at \((2, 0)\). On \([0, 2]\), the function is nonnegative. 2. The enclosed area is \(\int_0^2(x^3-12x+16)\, dx=\left[\frac{x^4}{4}-6x^2+16x\right]_0^2=12\). 3. The rectangle area is \(A(u)=u f(u)=u^4-12u^2+16u\). Differentiate: \(A'(u)=4u^3-24u+16=4(u-2)(u^2+2u-2)\). 4. The only critical point in \((0, 2)\) is \(u=\sqrt{3}-1\). Since \(A''(u)=12u^2-24<0\) at this value, and the area is zero at both boundary values, it gives the absolute maximum. 5. The y-coordinate is \(f(\sqrt{3}-1)=18-6\sqrt{3}\).

a) The enclosed area is \(12\) square units. b) The maximizing point is \(P(\sqrt{3}-1, 18-6\sqrt{3})\approx(0.732, 7.608)\).

- Use \(2x\) for the base length. - The height is \(10-f(x)\). - Substitute \(z=x^2\) in the critical-point equation. - Evaluate the area at the maximizing value.

1. The horizontal base has length \(2x\), and the vertical height is \(10-f(x)\). Thus \(A(x)=\frac{1}{2}(2x)(10-f(x))=-x^5+6x^3+8x\). The condition \(f(x)\le10\) gives \(0<x\le\sqrt{3+\sqrt{17}}\). 2. Differentiate: \(A'(x)=-5x^4+18x^2+8\). Let \(z=x^2\). The critical-point equation becomes \(5z^2-18z-8=0\). Its positive solution is \(z=4\), so \(x=2\). 3. The second derivative satisfies \(A''(2)=-88<0\). The area approaches zero as \(x\to0^+\) and equals zero at \(x=\sqrt{3+\sqrt{17}}\), so \(x=2\) gives the absolute maximum. 4. The maximum area is \(A(2)=-32+48+16=32\).

The area is maximized when \(x=2\), and the maximum area is \(32\) square units.

- When one object is above another, subtract the lower altitude from the higher altitude. - Simplify the difference before differentiating. - Find and classify the critical times. - Compare the critical values with both endpoint values.

1. Subtract the altitude functions: \(d(t) = -t^3 + 9t^2 - 5t + 60 - (10t + 50) = -t^3 + 9t^2 - 15t + 10\). 2. Differentiate: \(d'(t) = -3t^2 + 18t - 15 = -3(t - 1)(t - 5)\). The critical times are \(t = 1\) and \(t = 5\). 3. Use \(d''(t) = -6t + 18\). Since \(d''(1) = 12 > 0\), \(t = 1\) is a local minimum. Since \(d''(5) = -12 < 0\), \(t = 5\) is a local maximum. 4. Evaluate the candidates: \(d(0) = 10\), \(d(1) = 3\), \(d(5) = 35\), and \(d(7) = 3\). 5. The greatest separation is \(35\,\text{m}\), reached after \(5\) minutes.

a) \(d(t) = -t^3 + 9t^2 - 15t + 10\) b) The maximum separation is \(35\,\text{m}\) at \(t = 5\,\text{min}\).

- Intersections satisfy \(f(x)=g(x)\). - Write the vertical distance as \(f(x)-g(x)\) and maximize it. - The derivative gives each tangent slope. - Relate each tangent slope to its direction angle. - Use the angle-between-lines formula for two known slopes.

1. Solving \(f(x)=g(x)\) gives \(\frac{7}{20}x^2-\frac{21}{10}x=0\), so \(x=0\) or \(x=6\). Both functions equal \(0\) there, giving tips \((0, 0)\) and \((6, 0)\). 2. The vertical width is \(h(x)=f(x)-g(x)=-\frac{7}{20}x^2+\frac{21}{10}x\). Since \(h'(x)=-\frac{7}{10}x+\frac{21}{10}\), its critical value is \(x=3\). The second derivative is negative, and \(h(3)=\frac{63}{20}=3.15\). The maximum width is \(3.15\,\text{cm}\) at \(x=3\,\text{cm}\). 3. At \(x=6\), the tangent slopes are \(f'(6)=-\frac{3}{2}\) and \(g'(6)=\frac{3}{5}\). Thus, \(\tan\theta=\left|\frac{-\frac{3}{2}-\frac{3}{5}}{1+(-\frac{3}{2})(\frac{3}{5})}\right|=21\). Therefore, \(\theta=\arctan(21)\approx87.3^\circ\).

a) \((0, 0)\) and \((6, 0)\) b) \(3.15\,\text{cm}\) at \(x=3\,\text{cm}\) c) \(87.3^\circ\)

- Translate each endpoint and horizontal-tangent statement into an equation. - A horizontal tangent means the first derivative is \(0\). - Use the second derivative to find the inflection point. - Convert slope magnitude to percent and use inverse tangent for the angle.

1. The equation \(f(10)=0\) means the slide reaches ground level after \(10\,\text{m}\) horizontally. The equation \(f'(0)=0\) means the slide begins horizontally. 2. Let \(f(x)=ax^3+bx^2+cx+d\). The conditions \(f(0)=8\), \(f'(0)=0\), \(f(10)=0\), and \(f'(10)=0\) give \(d=8\), \(c=0\), \(a=\frac{2}{125}\), and \(b=-\frac{6}{25}\). Thus, \(f(x)=\frac{2}{125}x^3-\frac{6}{25}x^2+8\). 3. Since \(f''(x)=\frac{12}{125}x-\frac{12}{25}\), the inflection point occurs at \(x=5\), where \(f(5)=4\). Thus, \(W=(5, 4)\). The slide is steepest there and changes concavity. 4. The slope at \(x=5\) is \(f'(5)=-\frac{6}{5}\). Therefore, the maximum downward grade is \(120\%\), and the angle of descent is \(\arctan(\frac{6}{5})\approx50.2^\circ\).

a) The slide reaches the ground at \(x=10\) and begins horizontally at \(x=0\). b) \(f(x)=\frac{2}{125}x^3-\frac{6}{25}x^2+8\) c) \(W=(5, 4)\); this is the steepest point and where concavity changes. d) \(120\%\); \(50.2^\circ\)

- Smooth connections to horizontal platforms require matching heights and zero slopes. - Begin with a general cubic polynomial and its derivative. - Use the four endpoint conditions to determine the cubic. - The extremum of the slope occurs where the second derivative is \(0\). - Compare the maximum derivative value with \(0.6\).

1. Smooth horizontal connections give \(f(0)=0\), \(f'(0)=0\), \(f(4)=2\), and \(f'(4)=0\). Solving for a cubic gives \(f(x)=-\frac{1}{16}x^3+\frac{3}{8}x^2\). 2. The second derivative is \(f''(x)=-\frac{3}{8}x+\frac{3}{4}\), so the inflection point occurs at \(x=2\). Since \(f(2)=1\), the point is \(W=(2, 1)\). This is where the ramp is steepest. 3. The maximum slope is \(f'(2)=\frac{3}{4}=0.75\), or \(75\%\). Since \(75\%>60\%\), the design does not meet the requirement.

a) \(f(x)=-\frac{1}{16}x^3+\frac{3}{8}x^2\) b) \(W=(2, 1)\); the ramp is steepest there. c) No. The maximum grade is \(75\%\), which exceeds \(60\%\).

- Express the vertical difference between the curve and the line as a function. - Set the derivative of the harvest function equal to zero. - Convert from hundreds of deer to individual deer. - What does \(f(x) = x\) mean from one year to the next?

1. Write the harvest function: \(E(x) = -0.12x^2 + 1.44x\). 2. Differentiate: \(E'(x) = -0.24x + 1.44\). Set \(E'(x) = 0\): \(x = 6\). 3. Since \(E''(x) = -0.24 < 0\), this gives the absolute maximum. The optimal population is \(6\) hundred deer, or \(600\) deer. 4. Evaluate \(E(6) = 4.32\), which represents \(432\) deer per year. 5. The positive intersection satisfies \(f(x) = x\). Solving gives \(x = 12\), so a population of \(1200\) deer is an equilibrium: without harvesting or other changes, the model predicts the same population one year later.

a) \(600\) deer b) \(432\) deer per year c) The intersection at \(x = 12\) represents an equilibrium population of \(1200\) deer.

- The vertical intercept of a cost graph represents fixed cost. - Profit is the vertical distance between the revenue line and the cost curve. - Use the derivative of the profit function to find its maximum. - Convert hundreds of dollars to dollars in the final answer.

1. The fixed cost is \(C(0)=10\) hundreds of dollars, or \(\$1000\). 2. At \(x=8\), \(C(8)=0.2(8)^3-2(8)^2+8(8)+10=48.4\) hundreds of dollars, or \(\$4840\). 3. The revenue graph is the line \(R(x)=8x\), passing through \((0,0)\) and \((10,80)\). Its intersections with the cost graph occur at approximately \(x=2.60\) and \(x=9.44\), the break-even production levels. 4. Profit, in hundreds of dollars, is \(P(x)=R(x)-C(x)=-0.2x^3+2x^2-10\). 5. Differentiate: \(P'(x)=-0.6x^2+4x=-0.2x(3x-20)\). The relevant interior critical value is \(x=\frac{20}{3}\approx 6.67\). 6. The derivative changes from positive to negative there, so profit is maximized at \(x=\frac{20}{3}\). 7. The maximum profit is \(P\left(\frac{20}{3}\right)=\frac{530}{27}\approx 19.63\) hundreds of dollars, or approximately \(\$1962.96\).

a) The fixed cost is \(\$1000\), and the cost of producing \(8\) units is \(\$4840\). b) The break-even production levels are approximately \(x=2.60\) and \(x=9.44\). c) Profit is maximized at \(x_{\text{opt}}=\frac{20}{3}\approx 6.67\) units, with a maximum profit of approximately \(\$1962.96\).

- Use the first derivative to locate a maximum of the height function. - The steepest descent is the minimum value of the first derivative. - Locate an extremum of the first derivative by setting the second derivative equal to zero. - Apply both the product rule and the chain rule when differentiating.

1. Differentiate: \(f^{\prime}(x)=(-0.5x^2+2)e^{-0.5x}\) and \(f^{\prime\prime}(x)=(0.25x^2-x-1)e^{-0.5x}\). 2. Critical points of \(f\) satisfy \(-0.5x^2+2=0\), so \(x=\pm2\). On the given domain, \(x=2\) gives the maximum. Its height is \(f(2)=16e^{-1}\approx5.89\,\text{m}\). 3. The steepest descent occurs where \(f^{\prime}\) has a minimum, so set \(f^{\prime\prime}(x)=0\). This gives \(x^2-4x-4=0\), with solutions \(x=2\pm2\sqrt{2}\). 4. On the right side, \(x=2+2\sqrt{2}\approx4.83\,\text{m}\). The slope there is approximately \(f^{\prime}(4.83)\approx-0.864\). 5. The downward grade has magnitude \(0.864\), or approximately \(86.4\%\).

a) The maximum height is \(\frac{16}{e}\,\text{m}\approx5.89\,\text{m}\). b) The steepest descent occurs at \(x=2+2\sqrt{2}\approx4.83\,\text{m}\). The slope is approximately \(-0.864\), corresponding to a downward grade of about \(86.4\%\).

- Express the width and height of the rectangle in terms of \(u\). - Use area equals width times height. - Differentiate the area function using the product and chain rules. - Use the sign of the derivative and the end behavior to confirm an absolute maximum.

1. The rectangle has width \(2u\) and height \(g(u)=8e^{-u^2}\). Therefore, \(A(u)=2u\cdot8e^{-u^2}=16ue^{-u^2}\). 2. Using the product and chain rules, \(A^{\prime}(u)=16e^{-u^2}(1-2u^2)\). 3. Since \(e^{-u^2}>0\), \(A^{\prime}(u)=0\) for positive \(u\) only when \(u=\frac{1}{\sqrt{2}}\). 4. The derivative is positive for \(0<u<\frac{1}{\sqrt{2}}\) and negative for \(u>\frac{1}{\sqrt{2}}\). Also, \(A(u)\to0\) as \(u\to0^+\) and as \(u\to\infty\), so this critical point gives the unique absolute maximum. 5. The maximum area is \(A\left(\frac{1}{\sqrt{2}}\right)=8\sqrt{2}e^{-1/2}\approx6.86\) square units.

a) \(A(u)=16ue^{-u^2}\) b) \(u=\frac{1}{\sqrt{2}}\approx0.71\) c) \(A_{\max}=8\sqrt{2}e^{-1/2}\approx6.86\) square units

- Determine where the square root is defined and where the physical model begins. - Reflecting across the x-axis changes the sign of every output. - Use the derivative to find the tangent slope, then relate slope to angle. - First maximize the upper boundary; symmetry gives the full width.

1. a) The radicand must satisfy \(9-x\ge0\), so \(x\le9\). The modeled hull begins at the intercept \(x=-3\). Therefore, \(D_f=[-3,9]\). 2. b) Reflection across the x-axis changes each output to its opposite, so \(g(x)=-f(x)=-0.2(x+3)\sqrt{9-x}\). 3. c) By the product and chain rules, \(f'(x)=0.1\frac{15-3x}{\sqrt{9-x}}\). Thus, \(f'(-3)=\frac{2.4}{\sqrt{12}}=\frac{2\sqrt{3}}{5}\approx0.6928\). 4. The crossing angle is \(\alpha=\arctan\!\left(\frac{2\sqrt{3}}{5}\right)\approx34.7^\circ\). 5. d) The upper boundary is largest where \(f'(x)=0\), which gives \(15-3x=0\), so \(x=5\). The endpoint values are \(0\), while \(f(5)=0.2(8)(2)=3.2\,\text{cm}\). 6. By symmetry, the maximum width is \(2f(5)=6.4\,\text{cm}\).

a) \(D_f=[-3,9]\) b) \(g(x)=-0.2(x+3)\sqrt{9-x}\) c) \(\alpha\approx34.7^\circ\) d) \(6.4\,\text{cm}\)

- Write the width and height of the rectangle in terms of \(u\). - Substitute the function value for the height. - Differentiate the area function using the product rule. - Use the sign of the derivative to confirm that the critical point gives a maximum.

1. The rectangle has width \(u\) and height \(f(u)\), so \(A(u)=u\cdot f(u)=12u^2e^{-0.5u}\). 2. Using the product and chain rules, \(A^{\prime}(u)=(24u-6u^2)e^{-0.5u}=6u(4-u)e^{-0.5u}\). 3. For \(u>0\), the only critical point is \(u=4\). The derivative is positive on \((0, 4)\) and negative on \((4, \infty)\). 4. Also, \(A(u)\to0\) as \(u\to0^+\) and as \(u\to\infty\), so \(u=4\) gives the unique absolute maximum. 5. The maximum area is \(A(4)=192e^{-2}\approx25.98\) square units.

The area is maximized when \(u=4\). The maximum area is \(192e^{-2}\approx25.98\) square units.

- When every function value is negative, is the point closest to the x-axis at a maximum or a minimum? - Examine every critical number; not every horizontal tangent is an extremum. - Use the end behavior to decide whether the local extremum is absolute.

1. Since \(f(x) < 0\), the point closest to the x-axis occurs where \(f\) has its greatest value. 2. Differentiate: \(f'(x) = -x^3 + 3x^2 = -x^2(x - 3)\). The critical numbers are \(x = 0\) and \(x = 3\). 3. At \(x = 0\), the derivative does not change sign, so this critical point is not an extremum. At \(x = 3\), the derivative changes from positive to negative, so \(f\) has a local maximum. 4. Because \(f(x) \to -\infty\) as \(x \to \pm\infty\), this local maximum is also the absolute maximum. 5. Evaluate \(f(3) = -\frac{1}{4}(3^4) + 3^3 - 10 = -\frac{13}{4} = -3.25\). The closest point is \((3, -3.25)\).

\((3, -3.25)\)

- What condition must the first derivative satisfy at an interior local extremum? - What does the sign of the first derivative tell you about total cost? - Profit equals revenue minus cost. - Keep the units in thousands of dollars consistent. - Use derivatives and endpoint comparisons to find the absolute maximum.

1. Differentiate the cost function: \(K'(x) = x^2 - 8x + 20\). 2. The discriminant of \(x^2 - 8x + 20 = 0\) is \((-8)^2 - 4(1)(20) = -16\), so \(K'(x)\) has no real zeros. Since \(K'(0) = 20 > 0\), the cost function is increasing everywhere and has no local extrema. This is reasonable because total cost generally rises as production increases. 3. Revenue is \(R(x) = 40x\), so \(P(x) = R(x) - K(x) = -\frac{1}{3}x^3 + 4x^2 + 20x - 50\). 4. Differentiate: \(P'(x) = -x^2 + 8x + 20\). Solving \(P'(x) = 0\) gives \(x = -2\) and \(x = 10\). Only \(x = 10\) is in the production interval. 5. Since \(P''(x) = -2x + 8\) and \(P''(10) = -12 < 0\), \(x = 10\) gives a local maximum. 6. Compare values: \(P(0) = -50\), \(P(10) = \frac{650}{3}\), and \(P(15) = 25\). Therefore, the maximum profit is \(\frac{650}{3}\) thousand dollars, or approximately \(\$216{,}666.67\), when \(1000\) units are produced.

a) \(K'(x) = x^2 - 8x + 20\) has no real zeros, so \(K\) has no local extrema and is increasing throughout the interval. b) \(P(x) = -\frac{1}{3}x^3 + 4x^2 + 20x - 50\). The maximum profit is approximately \(\$216{,}666.67\), achieved by producing \(1000\) units.

- Solve each quadratic equation for its zeros. - Compare the x-intercepts of the two graphs. - Recall how even powers of \(x\) relate to y-axis symmetry. - Determine which graph is above the other between the intersections before setting up the area integral. - Differentiate the area function with respect to \(k\).

1. Solve \(f_k(x)=0\): \(kx^2-k^3=0\). Since \(k>0\), \(x^2=k^2\), so the zeros are \(x=\pm k\). 2. Similarly, \(g_k(x)=0\) gives \(\frac{1}{k}x^2-k=0\), so \(x^2=k^2\) and \(x=\pm k\). Thus, both graphs have x-intercepts \((-k, 0)\) and \((k, 0)\). 3. Each function contains only an \(x^2\) term and a constant term, so \(f_k(-x)=f_k(x)\) and \(g_k(-x)=g_k(x)\). Therefore, both graphs are symmetric about the y-axis. 4. For \(0<k<1\), \(f_k\) lies above \(g_k\) between \(-k\) and \(k\). By symmetry, \(A(k)=2\int_0^k\left[(k-\frac1k)x^2+k-k^3\right]\, \mathrm{d}x\) \(=\frac43(k^2-k^4)\). 5. Differentiate: \(A'(k)=\frac83k-\frac{16}{3}k^3\) \(=\frac83k(1-2k^2)\). The critical point in \((0, 1)\) is \(k=\frac{1}{\sqrt2}\). Also, \(A''(k)=\frac83-16k^2\), so \(A''\left(\frac{1}{\sqrt2}\right)=-\frac{16}{3}<0\). Therefore, \(A(k)\) has a local maximum there. The maximum area is \(A\left(\frac{1}{\sqrt2}\right)=\frac43\left(\frac12-\frac14\right)=\frac13\).

a) \(x=\pm k\) b) Both graphs have x-intercepts \((-k, 0)\) and \((k, 0)\). c) Both functions are even, so both graphs are symmetric about the y-axis. d) \(A(k)\) has a local maximum at \(k=\frac{1}{\sqrt2}\), and the maximum area is \(\frac13\) square unit.

- Factor out \(x\) to find each function's zeros. - Factor the difference function to analyze its sign. - Integrate the upper function minus the lower function over \([0, a]\). - Differentiate the resulting area function with respect to \(a\).

1. Factor each function: \(h_a(x)=x(a-x)\), so its zeros are \(x=0\) and \(x=a\). Also, \(p_a(x)=\frac{x}{a}\left(1-\frac{x}{a}\right)\), so its zeros are again \(x=0\) and \(x=a\). 2. The difference is \(d_a(x)=\left(\frac1a-a\right)x-\left(\frac1{a^2}-1\right)x^2\) \(=\frac{1-a^2}{a^2}x(a-x)\). For \(0<a<1\) and \(0<x<a\), every factor in this expression is positive. Therefore, \(d_a(x)>0\). 3. The enclosed area is \(A(a)=\int_0^a d_a(x)\, \mathrm{d}x\) \(=\frac{1-a^2}{a^2}\int_0^a x(a-x)\, \mathrm{d}x\) \(=\frac16a(1-a^2)=\frac16(a-a^3)\). 4. Differentiate: \(A'(a)=\frac16(1-3a^2)\). The critical point in \((0, 1)\) satisfies \(a^2=\frac13\), so \(a=\frac1{\sqrt3}\). Since \(A''(a)=-a<0\) for \(a>0\), this point is a local maximum. In addition, \(A(a)\to0\) as \(a\to0^+\) and as \(a\to1^-\), so it is the absolute maximum.

a) Both functions have zeros \(x=0\) and \(x=a\). b) \(d_a(x)=\frac{1-a^2}{a^2}x(a-x)>0\) for \(0<x<a\). c) The enclosed area is maximized when \(a=\frac1{\sqrt3}=\frac{\sqrt3}{3}\).

- Find where the graph crosses the x-axis. - Split the area integral where the function changes sign. - Write the total area as a function of \(a\). - Find the critical point and compare it with the endpoint values.

1. The zeros of \(f_a\) are \(x(x-a)=0\), so \(x=0\) and \(x=a\). Because \(a\in[0, 2]\), both zeros lie in the interval. 2. The graph is below the x-axis on \((0, a)\) and above it on \((a, 2)\). Therefore, \(A(a)=\int_0^a(ax-x^2)\, \mathrm{d}x+\int_a^2(x^2-ax)\, \mathrm{d}x\). Evaluating and simplifying gives \(A(a)=\frac13a^3-2a+\frac83\). 3. Differentiate: \(A'(a)=a^2-2\). The critical point in \([0, 2]\) is \(a=\sqrt2\). Since \(A''(a)=2a\) and \(A''(\sqrt2)>0\), this point is a local minimum. 4. Check the endpoints: \(A(0)=\frac83\), \(A(2)=\frac43\), and \(A(\sqrt2)=\frac{8-4\sqrt2}{3}\approx0.78\). Therefore, the absolute minimum occurs at \(a=\sqrt2\).

The total area is minimized when \(a=\sqrt2\). The minimum area is \(\frac{8-4\sqrt2}{3}\approx0.78\) square unit.

- Find the x-coordinate where \(y=c\) intersects \(y=\sqrt{x}\). - Split the area integral at the intersection point. - Express the total area as a function of \(c\). - Differentiate and compare the critical value with the endpoints.

1. The line and the graph intersect where \(\sqrt{x}=c\), so \(x=c^2\). 2. The total area is \(A(c)=\int_0^{c^2}(c-\sqrt{x})\, \mathrm{d}x+\int_{c^2}^1(\sqrt{x}-c)\, \mathrm{d}x\). Evaluating and simplifying gives \(A(c)=\frac23c^3-c+\frac23\). 3. Differentiate: \(A'(c)=2c^2-1\). The critical point in \([0, 1]\) is \(c=\frac1{\sqrt2}\). Since \(A''(c)=4c>0\) at this point, it is a local minimum. 4. The minimum value is \(A\left(\frac1{\sqrt2}\right)=\frac{2-\sqrt2}{3}\approx0.195\). The endpoint values are \(A(0)=\frac23\) and \(A(1)=\frac13\), so this critical point gives the absolute minimum.

The total area is minimized when \(c=\frac1{\sqrt2}=\frac{\sqrt2}{2}\). The minimum area is \(\frac{2-\sqrt2}{3}\approx0.195\) square unit.

- First express the volume as a function of the parameter \(k\). - Treat \(k\) as a constant while integrating with respect to \(x\). - Find critical values of the volume function. - Use the sign of the derivative to verify a maximum.

1. By the disc method, \(V(k)=\pi\int_1^k(f_k(x))^2\,\text{d}x=\pi\int_1^k\frac{1}{kx}\,\text{d}x=\frac{\pi\ln k}{k}\). 2. \(V'(k)=\frac{\pi(1-\ln k)}{k^2}\). The only critical value for \(k>1\) satisfies \(1-\ln k=0\), so \(k=e\). 3. For \(1<k<e\), \(V'(k)>0\), and for \(k>e\), \(V'(k)<0\). Thus, \(V\) increases and then decreases, so the global maximum occurs at \(k=e\).

The volume is maximized when \(k=e\).

- Express the volume in terms of \(a\) after evaluating the integral. - Find the critical values of the volume function. - Use a sign chart for the derivative to classify the extremum.

1. \(V(a)=\pi\int_a^{a^2}\frac{1}{x^2}\,\text{d}x=\pi[-\frac{1}{x}]_a^{a^2}=\pi\left(\frac{1}{a}-\frac{1}{a^2}\right)\). 2. \(V'(a)=\frac{\pi(2-a)}{a^3}\). The only critical value for \(a>1\) is \(a=2\). 3. The derivative is positive for \(1<a<2\) and negative for \(a>2\). Therefore, \(V\) has a global maximum at \(a=2\).

The volume has a global maximum at \(a=2\).

- Treat \(k\) as a constant when differentiating with respect to \(x\). - Substitute the critical x-value into the original function. - Then treat the minimum y-coordinate as a new function of \(k\). - Round only the final parameter value.

1. Differentiating with respect to \(x\) gives \(f_k'(x)=k-\frac{1}{kx^2}\) and \(f_k''(x)=\frac{2}{kx^3}>0\). 2. Setting the first derivative equal to zero gives \(x^2=\frac{1}{k^2}\). Since \(x>0\), \(x=\frac{1}{k}\), and this point is a local minimum. 3. The y-coordinate is \(f_k\left(\frac{1}{k}\right)=k^2+\frac{110}{k}+2\). Thus the minimum point is \(\left(\frac{1}{k}, k^2+\frac{110}{k}+2\right)\). 4. Minimize \(H(k)=k^2+\frac{110}{k}+2\) for \(k>0\). Then \(H'(k)=2k-\frac{110}{k^2}\). Setting this equal to zero gives \(2k^3=110\), so \(k=\sqrt[3]{55}\approx3.80\). Since \(H''(k)=2+\frac{220}{k^3}>0\), this value gives the minimum.

Minimum point: \(\left(\frac{1}{k}, k^2+\frac{110}{k}+2\right)\) Optimal parameter: \(k=\sqrt[3]{55}\approx3.80\)

- Differentiate with respect to \(x\) first. - Substitute the critical x-value into the original function. - Minimize the resulting y-coordinate as a function of \(a\). - Use the second derivative to verify the minimum.

1. Differentiating with respect to \(x\) gives \(g_a'(x)=a-\frac{a^3}{x^2}\) and \(g_a''(x)=\frac{2a^3}{x^3}>0\). 2. The critical-point equation gives \(x^2=a^2\). Since \(x>0\) and \(a>0\), \(x=a\), and this point is a local minimum. 3. Its y-coordinate is \(g_a(a)=2a^2+\frac{5}{a}\). Thus the minimum point is \(\left(a, 2a^2+\frac{5}{a}\right)\). 4. Minimize \(H(a)=2a^2+\frac{5}{a}\). Then \(H'(a)=4a-\frac{5}{a^2}\). Setting this equal to zero gives \(4a^3=5\), so \(a=\sqrt[3]{\frac{5}{4}}\approx1.08\). Since \(H''(a)=4+\frac{10}{a^3}>0\), this value gives the minimum.

Minimum point: \(\left(a, 2a^2+\frac{5}{a}\right)\) Optimal parameter: \(a=\sqrt[3]{\frac{5}{4}}\approx1.08\)

- Remember that the tank has no top when writing its surface-area formula. - Use the fixed volume to express one dimension in terms of the other. - Find and verify the minimum of the resulting one-variable area function. - Use the critical-point equation to compare the depth directly with the radius.

1. The volume constraint is \(\pi r^2h=50\), so \(h=\frac{50}{\pi r^2}\), with \(r>0\). 2. Because the tank has no top, its interior area is \(A=\pi r^2+2\pi rh\). Substituting for \(h\) gives \(A(r)=\pi r^2+2\pi r\left(\frac{50}{\pi r^2}\right)=\pi r^2+\frac{100}{r}\). 3. Differentiate: \(A'(r)=2\pi r-\frac{100}{r^2}\). Set the derivative equal to zero: \(2\pi r^3=100\Rightarrow r=\sqrt[3]{\frac{50}{\pi}}\approx 2.515\,\text{m}\). 4. Since \(A''(r)=2\pi+\frac{200}{r^3}>0\) for \(r>0\), the critical point is a local minimum. Also, \(A(r)\to\infty\) as \(r\to 0^+\) or as \(r\to\infty\), so it is the absolute minimum. 5. From the critical-point equation, \(\pi r^3=50\). Therefore, \(h=\frac{50}{\pi r^2}=r\approx 2.515\,\text{m}\). 6. The optimal ratio is \(h:r=1:1\). For a closed cylinder with fixed volume, the corresponding optimum satisfies \(h=2r\), so its ratio is \(2:1\).

a) \(A(r)=\pi r^2+\frac{100}{r}\), for \(r>0\). b) \(r\approx 2.515\,\text{m}\) and \(h\approx 2.515\,\text{m}\). c) The optimal ratio is \(h:r=1:1\). For a closed cylinder, the optimal ratio is \(2:1\).

- Write formulas for the surface area and volume of a rectangular box with a square base. - Use the fixed surface area to express the height in terms of the base side length. - Create a one-variable volume function. - Find the largest value of that function. - Compare the maximum with \(80\,\text{cm}^3\) and use continuity to justify your conclusion.

1. Let \(x\) be the side length of the square base and \(h\) be the height. The surface-area constraint is \(2x^2+4xh=120\). 2. Solve for the height: \(h=\frac{120-2x^2}{4x}=\frac{30}{x}-\frac{x}{2}\). For positive dimensions, \(0<x<\sqrt{60}\). 3. The volume is \(V(x)=x^2h=x^2\left(\frac{30}{x}-\frac{x}{2}\right)=30x-\frac{1}{2}x^3\). 4. Differentiate: \(V'(x)=30-\frac{3}{2}x^2\). Setting \(V'(x)=0\) gives \(x^2=20\), so \(x=\sqrt{20}\). Since \(V''(x)=-3x<0\) for \(x>0\), this critical point is a local maximum. 5. The maximum volume is \(V(\sqrt{20})=20\sqrt{20}=40\sqrt{5}\approx 89.44\,\text{cm}^3\). Also, \(V(x)\to 0\) as \(x\to 0^+\) and as \(x\to\sqrt{60}^{\, -}\), so this is the absolute maximum. 6. The function \(V\) is continuous on \((0, \sqrt{60})\). Because its values rise from near \(0\) to about \(89.44\) and then return toward \(0\), the Intermediate Value Theorem guarantees at least one box with volume exactly \(80\,\text{cm}^3\). Therefore, the requested box is possible.

Yes. The greatest possible volume with the given surface area is \(40\sqrt{5}\approx 89.44\,\text{cm}^3\), which is greater than \(80\,\text{cm}^3\). Continuity then guarantees that a volume of exactly \(80\,\text{cm}^3\) occurs.

- Use the surface-area and volume formulas for a closed cylinder. - Express the height in terms of the radius using the fixed surface area. - Write the volume as a function of the radius alone. - Find the absolute maximum on the physically meaningful domain. - Compare that maximum with the target volume.

1. For a closed cylinder, \(2\pi r^2+2\pi rh=24\pi\). Dividing by \(2\pi\) gives \(r^2+rh=12\), so \(h=\frac{12-r^2}{r}\). Positive dimensions require \(0<r<\sqrt{12}\). 2. The volume as a function of \(r\) is \(V(r)=\pi r^2h=\pi r^2\left(\frac{12-r^2}{r}\right)=\pi(12r-r^3)\). 3. Differentiate: \(V'(r)=\pi(12-3r^2)\). Setting \(V'(r)=0\) gives \(r^2=4\), so \(r=2\,\text{m}\). Since \(V''(r)=-6\pi r<0\) for \(r>0\), this critical point is a local maximum. 4. Also, \(V(r)\to 0\) as \(r\to 0^+\) and as \(r\to\sqrt{12}^{\, -}\). Therefore, the critical point gives the absolute maximum: \(V(2)=\pi(12\cdot 2-2^3)=16\pi\,\text{m}^3\). 5. Because \(16\pi<20\pi\), the requested capacity cannot be reached.

No. The greatest possible volume is \(16\pi\,\text{m}^3\), which is less than \(20\pi\,\text{m}^3\).

- List the segments that make up the outside border. Do not include the shared edge between the rectangle and triangle. - Use the area formula for an equilateral triangle. - Express the rectangle’s height in terms of its width using the fixed total area. - Write the perimeter as a function of \(b\) alone and minimize it.

1. The shared edge between the rectangle and triangle is not part of the outside perimeter. The outside perimeter is \(P=b+2h+2b=3b+2h\). 2. The area constraint is \(bh+\frac{\sqrt{3}}{4}b^2=10\). Solving for \(h\) gives \(h=\frac{10}{b}-\frac{\sqrt{3}}{4}b\). 3. Substitute into the perimeter: \(P(b)=3b+2\left(\frac{10}{b}-\frac{\sqrt{3}}{4}b\right)=\left(3-\frac{\sqrt{3}}{2}\right)b+\frac{20}{b}\). 4. Differentiate: \(P'(b)=3-\frac{\sqrt{3}}{2}-\frac{20}{b^2}\). Setting \(P'(b)=0\) gives \(b^2=\frac{20}{3-\frac{\sqrt{3}}{2}}=\frac{40}{6-\sqrt{3}}\). Therefore, \(b=\sqrt{\frac{40}{6-\sqrt{3}}}\approx 3.06\,\text{cm}\). Since \(P''(b)=\frac{40}{b^3}>0\) for \(b>0\), this critical point gives the minimum perimeter.

The optimal width is \(b=\sqrt{\frac{40}{6-\sqrt{3}}}\approx 3.06\,\text{cm}\).

- Write formulas for the volume and for each part of the cylinder’s surface area. - Multiply each type of surface area by its corresponding price. - Use the fixed volume to eliminate the height. - Find and verify the minimum of the resulting one-variable cost function.

1. The volume constraint is \(\pi r^2h=2\), so \(h=\frac{2}{\pi r^2}\), with \(r>0\). 2. The top and bottom have total area \(2\pi r^2\), and the curved side has area \(2\pi rh\). Therefore, the cost is \(C(r, h)=10(2\pi r^2)+5(2\pi rh)=20\pi r^2+10\pi rh\). Substituting for \(h\) gives \(C(r)=20\pi r^2+\frac{20}{r}\). 3. Differentiate: \(C'(r)=40\pi r-\frac{20}{r^2}\). Set the derivative equal to zero: \(40\pi r^3=20\Rightarrow r^3=\frac{1}{2\pi}\). Thus \(r=\sqrt[3]{\frac{1}{2\pi}}\approx 0.542\,\text{m}\). 4. Since \(C''(r)=40\pi+\frac{40}{r^3}>0\) for \(r>0\), the critical point is a local minimum. Also, \(C(r)\to\infty\) as \(r\to 0^+\) or as \(r\to\infty\), so it is the absolute minimum. 5. The corresponding height is \(h=\frac{2}{\pi r^2}=2\sqrt[3]{\frac{4}{\pi}}\approx 2.168\,\text{m}\).

The minimum-cost dimensions are \(r=\sqrt[3]{\frac{1}{2\pi}}\approx 0.542\,\text{m}\) and \(h=2\sqrt[3]{\frac{4}{\pi}}\approx 2.168\,\text{m}\).

- Use the zeros and vertex of the parabola to determine the building’s width and height. - Apply the distance formula from \((0, 0)\) to \((x, f(x))\). - To minimize a square root, you may minimize its nonnegative radicand instead. - Recognize \(g(x)=\sqrt{64-x^2}\) as the upper half of a circle. - Divide the area difference by the stated reference area when computing the percent difference.

1. For part a, \(f(-8)=f(8)=0\), so the floor extends from \(x=-8\) to \(x=8\) and is \(16\,\text{m}\) wide. The parabola opens downward and has vertex \((0, 8)\), so its maximum height is \(8\,\text{m}\). 2. For part b, the distance formula gives \(d(x)=\sqrt{x^2+[f(x)]^2}\) \(=\sqrt{x^2+(8-0.125x^2)^2}\) \(=\sqrt{0.015625x^4-x^2+64}\). 3. For part c, minimizing \(d(x)\) is equivalent to minimizing its nonnegative radicand \(q(x)=0.015625x^4-x^2+64\). Differentiate: \(q'(x)=0.0625x^3-2x=x(0.0625x^2-2)\). The critical points are \(x=0\) and \(x=\pm\sqrt{32}=\pm 4\sqrt{2}\). The point \(x=0\) gives a local maximum of \(q\), while \(x=\pm 4\sqrt{2}\) give minima. The minimum distance is \(d(4\sqrt{2})=\sqrt{0.015625(32)^2-32+64}=\sqrt{48}=4\sqrt{3}\approx 6.93\,\text{m}\). 4. For part d, \(A_f=\int_{-8}^{8}(8-0.125x^2)\, dx=\frac{256}{3}\,\text{m}^2\). The graph of \(g\) is the upper semicircle of radius \(8\), so \(A_g=\frac{1}{2}\pi(8)^2=32\pi\,\text{m}^2\). The percent difference relative to \(A_g\) is \(\frac{A_g-A_f}{A_g}\cdot 100\%=\left(1-\frac{8}{3\pi}\right)100\%\approx 15.1\%\). Thus the parabolic model has about \(15.1\%\) less cross-sectional area.

a) The floor width is \(16\,\text{m}\), and the maximum height is \(8\,\text{m}\). b) \(d(x)=\sqrt{0.015625x^4-x^2+64}\). c) The wall is closest at \(x=\pm 4\sqrt{2}\approx\pm 5.66\). The minimum distance is \(4\sqrt{3}\approx 6.93\,\text{m}\). d) The area under \(f\) is about \(15.1\%\) less than the area under \(g\).

- Convert the clearance to the same unit used for the volume dimensions. - Because the tank is open, include one base and four side walls. - Express the total height as the water depth plus the required clearance. - Use the volume constraint to eliminate the water depth, then minimize the resulting area function.

1. Convert the clearance to meters: \(20\,\text{cm}=0.2\,\text{m}\). Let \(h_w\) be the water depth. The volume constraint is \(a^2h_w=2\), so \(h_w=\frac{2}{a^2}\). 2. The total height is \(H=\frac{2}{a^2}+0.2\). 3. Because the tank is open at the top, the material area is \(A=a^2+4aH\). Substituting for \(H\) gives \(A(a)=a^2+\frac{8}{a}+0.8a\), for \(a>0\). 4. Differentiate: \(A'(a)=2a-\frac{8}{a^2}+0.8\). The critical-point equation is \(a^3+0.4a^2-4=0\). Its positive solution is \(a\approx 1.465\,\text{m}\). 5. Since \(A''(a)=2+\frac{16}{a^3}>0\) for \(a>0\), the critical point gives the minimum. The corresponding total height is \(H=\frac{2}{(1.464644\ldots)^2}+0.2\approx 1.132\,\text{m}\).

The material is minimized when \(a\approx 1.465\,\text{m}\) and \(H\approx 1.132\,\text{m}\).

- Find each graph's local maximum in terms of \(k\), then eliminate the parameter. - For part b, hold \(x\) fixed and view the expression as a function of \(k\). - Find the greatest y-value the family can attain at that fixed x-coordinate.

1. Differentiate with respect to \(x\): \(g_k'(x)=e^{k-x}(1-x+k)\). The only critical number is \(x=k+1\). Since \(g_k''(k+1)=-e^{-1}<0\), the graph has a local maximum there. Its coordinates are \(\left(k+1, k+\frac1e\right)\). Since \(x=k+1\), we have \(k=x-1\), so the maxima lie on \(y=x-1+\frac1e\). 2. Fix \(x\) and vary \(k\). Let \(t=k-x\). Then \(g_k(x)=x+t-te^t\). As a function of \(t\), its derivative is \(1-(t+1)e^t\). This derivative is positive for \(t<0\), zero at \(t=0\), and negative for \(t>0\). Also, the expression approaches \(-\infty\) as \(t\to\pm\infty\). Therefore, its absolute maximum occurs at \(t=0\), or \(k=x\), and the maximum value is \(x\). Thus, for each fixed \(x\), the values attained by the family are exactly \((-\infty, x]\). A point lies on no graph exactly when \(y>x\).

a) \(y=x-1+\frac1e\) b) All points \((x, y)\) such that \(y>x\).

- Relate the width and height of the rectangular cross section to the diameter of the circular log. - Use the constraint to write the stiffness as a function of one variable. - Differentiate using the product and chain rules and compare the critical point with the endpoints. - For the general proof, keep the diameter as \(d\) rather than substituting \(40\).

1. A rectangle inscribed in a circle of diameter \(d\) has diagonal \(d\), so \(b^2+h^2=d^2\). Thus \(h^2=d^2-b^2\), and \(S(b)=kb(d^2-b^2)^{3/2}\), for \(0\le b\le d\). 2. Differentiate using the product and chain rules: \(S'(b)=k\sqrt{d^2-b^2}(d^2-4b^2)\). The interior critical point satisfies \(d^2-4b^2=0\Rightarrow b=\frac{d}{2}\). The stiffness is \(0\) at both endpoints of the domain and positive at this interior critical point, so it gives the absolute maximum. 3. For \(d=40\,\text{cm}\), \(b=20\,\text{cm}\) and \(h=\sqrt{40^2-20^2}=20\sqrt{3}\approx 34.64\,\text{cm}\). 4. In general, when \(b=\frac{d}{2}\), \(h^2=d^2-\frac{d^2}{4}=\frac{3d^2}{4}\), so \(h=\frac{\sqrt{3}}{2}d\). Therefore, \(\frac{h}{b}=\frac{\frac{\sqrt{3}}{2}d}{\frac{1}{2}d}=\sqrt{3}\). The claimed ratio is independent of \(d\).

a) \(b=20\,\text{cm}\) and \(h=20\sqrt{3}\approx 34.64\,\text{cm}\). b) For any diameter \(d\), the maximizing dimensions satisfy \(b=\frac{d}{2}\) and \(h=\frac{\sqrt{3}}{2}d\), so \(\frac{h}{b}=\sqrt{3}\).

- Use a cross section through the sphere’s center, the pyramid’s apex, and a diagonal of the square base. - Relate the base circumradius, the pyramid height, and the sphere radius with the Pythagorean theorem. - Express \(a^2\) in terms of \(h\) before substituting into the volume formula. - Compare the critical-point value with the endpoint cases.

1. The volume is \(V=\frac{1}{3}a^2h\). In a cross section through the pyramid axis and a diagonal of the base, the circumradius of the square base is \(r_b=\frac{a}{\sqrt{2}}\). The base plane is a distance \(|h-R|\) from the sphere’s center, so \(r_b^2+(h-R)^2=R^2\). 2. Therefore, \(\frac{a^2}{2}=R^2-(h-R)^2=2Rh-h^2\), and \(a^2=4Rh-2h^2\). 3. Substitute into the volume formula: \(V(h)=\frac{1}{3}(4Rh-2h^2)h=\frac{4}{3}Rh^2-\frac{2}{3}h^3\), for \(0\le h\le 2R\). 4. Differentiate: \(V'(h)=\frac{8}{3}Rh-2h^2=2h\left(\frac{4R}{3}-h\right)\). The nonzero critical point is \(h=\frac{4R}{3}\). The volume is \(0\) at \(h=0\) and \(h=2R\), so this interior critical point gives the absolute maximum. 5. Substituting the maximizing height into the constraint gives \(a^2=4R\left(\frac{4R}{3}\right)-2\left(\frac{4R}{3}\right)^2=\frac{16R^2}{9}\), so \(a=\frac{4R}{3}\).

The volume is maximized when \(h=\frac{4R}{3}\) and \(a=\frac{4R}{3}\).

- Use the two given relationships to express \(b\) and \(c\) in terms of \(a\). - For the area, maximizing its square avoids differentiating a square root. - When the triangle rotates about \(a\), identify which leg becomes the cone height and which becomes the radius. - Check that each critical point lies in the physical domain.

1. The constraints are \(a+c=12\) and \(a^2+b^2=c^2\). Thus \(c=12-a\), and \(b^2=(12-a)^2-a^2=144-24a\), so the physical domain is \(0<a<6\). 2. For part a, \(A(a)=\frac{1}{2}a\sqrt{144-24a}\). Because \(A\ge 0\), maximizing \(A\) is equivalent to maximizing \(A(a)^2=36a^2-6a^3\). Differentiate: \(\frac{d}{da}[A(a)^2]=72a-18a^2=18a(4-a)\). The interior critical point is \(a=4\). The squared area approaches \(0\) at both ends of the domain, so this point gives the absolute maximum. Then \(c=8\,\text{cm}\) and \(b=\sqrt{144-96}=4\sqrt{3}\approx 6.93\,\text{cm}\). 3. For part b, rotation about \(a\) makes \(a\) the cone height and \(b\) the radius. Therefore, \(V(a)=\frac{1}{3}\pi b^2a=\frac{1}{3}\pi(144-24a)a=48\pi a-8\pi a^2\). Differentiate: \(V'(a)=48\pi-16\pi a\). The critical point is \(a=3\). Since the quadratic volume function is concave down, this is the absolute maximum. Then \(c=9\,\text{cm}\) and \(b=\sqrt{144-72}=6\sqrt{2}\approx 8.49\,\text{cm}\).

a) \(a=4\,\text{cm}\), \(b=4\sqrt{3}\approx 6.93\,\text{cm}\), and \(c=8\,\text{cm}\). b) \(a=3\,\text{cm}\), \(b=6\sqrt{2}\approx 8.49\,\text{cm}\), and \(c=9\,\text{cm}\).

- Draw an axial cross section to relate the cone’s base radius, height, and the sphere radius. - Use the Pythagorean theorem to express the base radius in terms of \(h\) and \(R\). - Substitute into the cone-volume formula and optimize on \([0, 2R]\). - Compare the maximum cone volume with the sphere-volume formula.

1. The cone volume is \(V=\frac{1}{3}\pi r^2h\). In an axial cross section, the base radius, sphere radius, and cone height satisfy \(r^2+(h-R)^2=R^2\). Therefore, \(r^2=2Rh-h^2\). 2. Substitute into the volume formula: \(V(h)=\frac{\pi}{3}(2Rh^2-h^3)\), for \(0\le h\le 2R\). 3. Differentiate: \(V'(h)=\frac{\pi}{3}(4Rh-3h^2)=\frac{\pi h}{3}(4R-3h)\). The interior critical point is \(h=\frac{4R}{3}\). The volume is \(0\) at both endpoints of the domain, so this critical point gives the absolute maximum. 4. At this height, \(V_{\max}=\frac{32}{81}\pi R^3\). The sphere volume is \(V_{\text{sphere}}=\frac{4}{3}\pi R^3\). Thus \(\frac{V_{\max}}{V_{\text{sphere}}}=\frac{8}{27}\).

The maximizing height is \(h=\frac{4R}{3}\). The maximum cone volume is \(\frac{8}{27}\) of the sphere’s volume.

- Use a right triangle in an axial cross section to relate \(r\), \(h\), and \(R\). - Write the lateral surface area as circumference times height. - Squaring the nonnegative objective function can remove the square root before differentiation. - Substitute the optimal height back into the geometric constraint to find the ratio.

1. The lateral surface area is \(M=2\pi rh\). In an axial cross section, the cylinder’s diagonal is the sphere’s diameter, so \((2r)^2+h^2=(2R)^2\). Thus \(r=\sqrt{R^2-\frac{h^2}{4}}\). 2. Therefore, \(M(h)=2\pi h\sqrt{R^2-\frac{h^2}{4}}\), for \(0\le h\le 2R\). Since \(M\ge 0\), maximizing \(M\) is equivalent to maximizing \(M(h)^2=\pi^2(4R^2h^2-h^4)\). Ignoring the positive constant \(\pi^2\), let \(q(h)=4R^2h^2-h^4\). 3. Differentiate: \(q'(h)=8R^2h-4h^3=4h(2R^2-h^2)\). The interior critical point is \(h=\sqrt{2}R\). The lateral area is \(0\) at both endpoints, so this critical point gives the absolute maximum. 4. The corresponding radius is \(r=\sqrt{R^2-\frac{2R^2}{4}}=\frac{R}{\sqrt{2}}\). Therefore, \(\frac{h}{r}=\frac{\sqrt{2}R}{R/\sqrt{2}}=2\).

The lateral surface area is maximized when \(h=\sqrt{2}R\). At that point, \(r=\frac{R}{\sqrt{2}}\), so \(\frac{h}{r}=2\).

- At equilibrium, the population does not change from one year to the next. - Write natural growth as a function of the current population. - Find the maximum of that quadratic function. - Compare the fixed harvest with the largest possible natural increase.

1. At equilibrium, \(B_{n+1} = B_n\), so the harvest must equal the natural annual growth: \(A = G(B) = 0.12B\left(1 - \frac{B}{12500}\right)\). 2. Differentiate: \(G'(B) = 0.12 - \frac{0.24}{12500}B\). 3. Set \(G'(B) = 0\): \(B = 6250\). Because \(G\) is a downward-opening quadratic, this value gives the maximum growth. 4. Evaluate \(G(6250) = 0.12(6250)\left(1 - \frac{6250}{12500}\right) = 375\). 5. Therefore, the greatest sustainable fixed harvest in this model is \(375\) trees per year, with equilibrium population \(6250\).

The maximum fixed harvest is \(375\) trees per year. At that harvest level, the equilibrium population is \(6250\) trees.

- Place the intact opposite corner at the origin and locate the endpoints of the break line. - Find the equation of the line through those endpoints. - Express the area of an axis-aligned rectangle whose upper-right corner lies on the break. - Compare the interior maximum with the cases at the ends of the break segment.

1. Place the origin at the intact corner opposite the break. Let the original sheet occupy \(0\le x\le 120\) and \(0\le y\le 80\). The break joins \((60, 80)\) and \((120, 40)\). 2. The break line has slope \(\frac{40-80}{120-60}=-\frac{2}{3}\), so its equation is \(y=-\frac{2}{3}x+120\), for \(60\le x\le 120\). 3. If the upper-right corner of the new rectangle lies on the break line, its area is \(A(x)=x\left(-\frac{2}{3}x+120\right)=-\frac{2}{3}x^2+120x\). Differentiate: \(A'(x)=-\frac{4}{3}x+120\). The critical point is \(x=90\,\text{cm}\). Since \(A''(x)=-\frac{4}{3}<0\), it is a maximum on the line segment. 4. The corresponding height is \(y=-\frac{2}{3}(90)+120=60\,\text{cm}\). For completeness, the largest rectangle whose right edge is before the break begins has area \(60\cdot 80=4800\,\text{cm}^2\), and the endpoint at \(x=120\) also gives \(4800\,\text{cm}^2\). The \(90\times 60\) rectangle has area \(5400\,\text{cm}^2\), so it is the absolute maximum.

The largest new glass rectangle measures \(90\,\text{cm}\times 60\,\text{cm}\).

- Use the volume and total surface-area formulas for a closed cylinder. - Treat the original surface area as the fixed constraint in part b. - Express the height in terms of the radius before maximizing volume. - Compare the two volumes using percent increase.

1. For part a, \(V_1=\pi(3.3)^2(11.5)\approx 393.44\,\text{cm}^3\). The total surface area is \(S=2\pi(3.3)^2+2\pi(3.3)(11.5)=97.68\pi\approx 306.87\,\text{cm}^2\). 2. For a can with this fixed surface area, \(S=2\pi r^2+2\pi rh\), so \(h=\frac{S}{2\pi r}-r\). The volume is therefore \(V(r)=\pi r^2h=\frac{S}{2}r-\pi r^3\). 3. Differentiate: \(V'(r)=\frac{S}{2}-3\pi r^2\). The positive critical point is \(r=\sqrt{\frac{S}{6\pi}}=\sqrt{16.28}\approx 4.035\,\text{cm}\). Since \(V''(r)=-6\pi r<0\) for \(r>0\), this point gives the maximum. The surface-area constraint gives \(h=2r\approx 8.070\,\text{cm}\). 4. The optimized volume is \(V_{\max}\approx 412.73\,\text{cm}^3\). The percent increase is \(\left(\frac{V_{\max}}{V_1}-1\right)100\%\approx 4.90\%\).

a) \(V\approx 393.44\,\text{cm}^3\) and \(S\approx 306.87\,\text{cm}^2\). b) \(r\approx 4.035\,\text{cm}\) and \(h\approx 8.070\,\text{cm}\). c) The optimized volume is about \(4.90\%\) greater.

- Derive the radius-height relationship using similar triangles. - Maximize the cylinder volume within a general cone before using the given dimensions. - Apply the same optimal proportions to the smaller remaining cone. - Add both cylinder volumes and divide by the original cone volume.

1. For a cylinder inscribed with its base in a cone’s base, similar triangles give \(r=R\left(1-\frac{h}{H}\right)\). Thus \(V(h)=\pi R^2h\left(1-\frac{h}{H}\right)^2\). Differentiating and factoring gives \(V'(h)=\pi R^2\left(1-\frac{h}{H}\right)\left(1-\frac{3h}{H}\right)\). On \(0<h<H\), the maximum occurs at \(h=\frac{H}{3}\), with \(r=\frac{2R}{3}\). 2. For the original cone, \(h_1=6\,\text{cm}\) and \(r_1=6\,\text{cm}\). The remaining cone has height \(12\,\text{cm}\) and radius \(6\,\text{cm}\). Therefore, the second cylinder has \(h_2=4\,\text{cm}\) and \(r_2=4\,\text{cm}\). 3. The original cone volume is \(V_K=\frac{1}{3}\pi(9)^2(18)=486\pi\,\text{cm}^3\). The cylinder volumes are \(V_1=\pi(6)^2(6)=216\pi\,\text{cm}^3\) and \(V_2=\pi(4)^2(4)=64\pi\,\text{cm}^3\). Hence \(\frac{V_1+V_2}{V_K}=\frac{280}{486}=\frac{140}{243}\approx 0.576\).

1. First cylinder: \(r_1=6\,\text{cm}\), \(h_1=6\,\text{cm}\). Second cylinder: \(r_2=4\,\text{cm}\), \(h_2=4\,\text{cm}\). 2. The volume ratio is \(\frac{140}{243}\approx 0.576\).

- Add the cylinder and cone volumes to form the constraint. - Use the Pythagorean theorem to find the cone’s slant height. - Add only the cylindrical lateral area and conical lateral area. - Eliminate \(h\), then minimize the resulting function of \(r\).

1. The volume constraint is \(\pi r^2h+\frac{1}{3}\pi r^3=600\), so \(h=\frac{600}{\pi r^2}-\frac{r}{3}\). 2. The cone’s slant height is \(s=\sqrt{r^2+r^2}=r\sqrt{2}\). The fabric area is \(A=2\pi rh+\pi rs\). Substituting for \(h\) and \(s\) gives \(A(r)=\frac{1200}{r}+\pi r^2\left(\sqrt{2}-\frac{2}{3}\right)\). 3. Differentiate: \(A'(r)=-\frac{1200}{r^2}+2\pi r\left(\sqrt{2}-\frac{2}{3}\right)\). The critical point satisfies \(r^3=\frac{600}{\pi\left(\sqrt{2}-\frac{2}{3}\right)}\), so \(r\approx 6.345\,\text{m}\). 4. Since \(A''(r)=\frac{2400}{r^3}+2\pi\left(\sqrt{2}-\frac{2}{3}\right)>0\), the critical point gives a minimum. The corresponding cylindrical height is \(h=\frac{600}{\pi(6.345331\ldots)^2}-\frac{6.345331\ldots}{3}\approx 2.628\,\text{m}\).

The fabric use is minimized when \(r\approx 6.345\,\text{m}\) and \(h\approx 2.628\,\text{m}\).

- Use an axial cross section to relate the cylinder dimensions to the hemisphere radius. - Express \(r^2\) in terms of \(h\). - Substitute into the cylinder-volume formula. - Find and verify the interior maximum.

1. In an axial cross section, the radius and height satisfy \(r^2+h^2=10^2\), so \(r^2=100-h^2\). 2. The cylinder volume is \(V(h)=\pi(100-h^2)h=100\pi h-\pi h^3\), for \(0\le h\le 10\). 3. Differentiate: \(V'(h)=100\pi-3\pi h^2\). The positive critical point is \(h=\frac{10}{\sqrt{3}}\approx 5.77\,\text{cm}\). Since \(V''(h)=-6\pi h<0\) for \(h>0\), this point gives the maximum. 4. The corresponding radius is \(r=\sqrt{100-\frac{100}{3}}=10\sqrt{\frac{2}{3}}\approx 8.16\,\text{cm}\).

The maximum-volume cylinder has \(h=\frac{10}{\sqrt{3}}\approx 5.77\,\text{cm}\) and \(r=10\sqrt{\frac{2}{3}}\approx 8.16\,\text{cm}\).

- For a fixed base size, use the largest permitted length. - Determine which length restriction is active for each range of \(a\). - Optimize the volume separately on the two ranges. - Compare the largest volumes from both cases.

1. The volume is \(V=a^2l\). For a fixed \(a\), the volume increases with \(l\), so choose \(l=\min(80, 160-4a)\). 2. When \(0<a\le 20\), the length cap is active and \(l=80\). Then \(V_1(a)=80a^2\), which is increasing. Its largest value on this interval occurs at \(a=20\) and is \(32{,}000\,\text{cm}^3\). 3. When \(20\le a<40\), the combined-size constraint is active and \(l=160-4a\). Then \(V_2(a)=a^2(160-4a)=160a^2-4a^3\). Differentiate: \(V_2'(a)=320a-12a^2=4a(80-3a)\). The interior critical point is \(a=\frac{80}{3}\,\text{cm}\). Since \(V_2''\left(\frac{80}{3}\right)=-320<0\), this point gives the maximum in the second interval. 4. The corresponding length is \(l=160-4\left(\frac{80}{3}\right)=\frac{160}{3}\,\text{cm}\). The volume is \(V=\left(\frac{80}{3}\right)^2\left(\frac{160}{3}\right)=\frac{1024000}{27}\approx 37{,}925.93\,\text{cm}^3\). This exceeds \(32{,}000\,\text{cm}^3\), so it is the global maximum.

The maximum occurs at \(a=\frac{80}{3}\approx 26.67\,\text{cm}\) and \(l=\frac{160}{3}\approx 53.33\,\text{cm}\). The maximum volume is \(\frac{1024000}{27}\approx 37{,}925.93\,\text{cm}^3\).

- Write an equation for the total length of the eight metal-edged segments. - Substitute the resulting expression for \(h\) into the volume formula. - Use the minimum-height condition to determine the feasible interval for \(a\). - Check whether the unconstrained critical point lies in that interval; otherwise inspect the boundary.

1. The edging constraint is \(4a+4h=360\), so \(h=90-a\). Therefore, \(V(a)=a^2h=a^2(90-a)=90a^2-a^3\). 2. The requirement \(h\ge 40\) gives \(90-a\ge 40\Rightarrow 0<a\le 50\). 3. Differentiate: \(V'(a)=180a-3a^2=3a(60-a)\). This derivative is positive throughout \((0, 50]\), because the unconstrained interior maximum would occur at \(a=60\), outside the allowed domain. Therefore, \(V\) is increasing on the entire feasible interval. 4. The constrained maximum occurs at the right endpoint: \(a=50\,\text{cm}\). Then \(h=90-50=40\,\text{cm}\), and \(V_{\max}=50^2(40)=100{,}000\,\text{cm}^3\).

a) \(V(a)=90a^2-a^3\). b) \(a=50\,\text{cm}\) and \(h=40\,\text{cm}\). c) \(V_{\max}=100{,}000\,\text{cm}^3\).

- Resolve each slanted side panel into horizontal and vertical components. - Use the trapezoid-area formula. - Differentiate with angles measured in radians. - Restrict the solution to physically meaningful angles and compare boundary behavior.

1. The trapezoid height is \(10\sin\alpha\). Its lower base is \(10\), and its upper base is \(10+20\cos\alpha\). Therefore, \(A(\alpha)=\frac{10+(10+20\cos\alpha)}{2}(10\sin\alpha)\) \(=100\sin\alpha(1+\cos\alpha)\), for \(0<\alpha\le 90^\circ\). 2. Differentiate with respect to \(\alpha\), measured in radians: \(A'(\alpha)=100(\cos\alpha+\cos 2\alpha)\). Setting the derivative equal to zero gives \(\cos\alpha+2\cos^2\alpha-1=0\). Let \(u=\cos\alpha\). Then \(2u^2+u-1=0\), so \(u=\frac{1}{2}\) or \(u=-1\). Only \(u=\frac{1}{2}\) is in the stated angle range, giving \(\alpha=60^\circ\). 3. Comparing with the endpoint behavior confirms the maximum. The area is \(A(60^\circ)=100\left(\frac{\sqrt{3}}{2}\right)\left(\frac{3}{2}\right)=75\sqrt{3}\approx 129.90\,\text{cm}^2\).

The optimal angle is \(\alpha=60^\circ\), and the maximum cross-sectional area is \(75\sqrt{3}\approx 129.90\,\text{cm}^2\).

- Express the four triangular face areas using the slant height. - Relate slant height, vertical height, and half the base edge. - Maximizing the square of the nonnegative volume removes a square root. - Combine the critical-point equation with the fixed-area constraint.

1. The slant height of each triangular face is \(s=\sqrt{H^2+\frac{a^2}{4}}\). The lateral area is \(M=4\left(\frac{1}{2}as\right)=a\sqrt{4H^2+a^2}\). Squaring and solving for \(H^2\) gives \(H^2=\frac{M^2-a^4}{4a^2}\). 2. The volume is \(V=\frac{1}{3}a^2H\). Because \(V\ge 0\), maximize \(V^2=\frac{1}{36}(M^2a^2-a^6)\), for \(0<a\le\sqrt{M}\). 3. Ignoring the positive constant, differentiate \(f(a)=M^2a^2-a^6\): \(f'(a)=2M^2a-6a^5\). The positive critical point satisfies \(M^2=3a^4\). At this point, \(f''(a)=2M^2-30a^4=-24a^4<0\), so it is a local maximum. Also, \(V\to 0\) as \(a\to 0^+\), and \(V=0\) when \(a=\sqrt{M}\), so the critical point gives the absolute maximum. 4. Combining the critical-point equation with \(M^2=4a^2H^2+a^4\) gives \(3a^4=4a^2H^2+a^4\Rightarrow a^2=2H^2\). Thus \(\frac{H}{a}=\frac{1}{\sqrt{2}}\).

The volume is maximized when \(H=\frac{a}{\sqrt{2}}\), so the optimal ratio is \(H:a=1:\sqrt{2}\).

- Because the tank is open, include only the base and the curved side. - Build the different material prices directly into the cost function. - Use the volume equation to eliminate \(h\). - Find where the derivative of the cost function is zero.

1. The material cost is \(K(r, h)=15\pi r^2+10(2\pi rh)\). The volume constraint is \(\pi r^2h=10\). 2. Solve for \(h\): \(h=\frac{10}{\pi r^2}\). 3. Substitute into the cost function: \(K(r)=15\pi r^2+20\pi r\left(\frac{10}{\pi r^2}\right)=15\pi r^2+\frac{200}{r}\). 4. Differentiate and set the derivative equal to zero: \(K'(r)=30\pi r-\frac{200}{r^2}=0\). Therefore, \(30\pi r^3=200\), so \(r=\sqrt[3]{\frac{20}{3\pi}}\approx 1.29\,\text{m}\). 5. Because \(K''(r)=30\pi+\frac{400}{r^3}>0\) for \(r>0\), and \(K(r)\to\infty\) as \(r\to 0^+\) or \(r\to\infty\), this critical point gives the absolute minimum. 6. The corresponding height is \(h=\frac{10}{\pi r^2}\approx 1.93\,\text{m}\). 7. The minimum cost is \(K\approx \$233.45\).

The minimum cost occurs at \(r\approx 1.29\,\text{m}\) and \(h\approx 1.93\,\text{m}\). The minimum material cost is \(K\approx \$233.45\).

- First express the cone's slant height in terms of \(r\). - Write equations for the exterior surface area and total volume. - Use the fixed surface area to eliminate \(h\). - Differentiate the resulting volume function to find its maximum.

1. The cone slant height is \(s=\sqrt{r^2+H^2}=\sqrt{r^2+\left(\frac{4}{3}r\right)^2}=\frac{5}{3}r\). 2. The exterior surface area is the circular base, the cylindrical side, and the cone's lateral surface: \(O=\pi r^2+2\pi rh+\pi rs=\frac{8}{3}\pi r^2+2\pi rh\). 3. The total volume is \(V=\pi r^2h+\frac{1}{3}\pi r^2H=\pi r^2h+\frac{4}{9}\pi r^3\). 4. Solve the surface-area equation for \(h\): \(h=\frac{O}{2\pi r}-\frac{4}{3}r\). 5. Substitute into the volume: \(V(r)=\frac{Or}{2}-\frac{8}{9}\pi r^3\). 6. Differentiate and set the derivative equal to zero: \(V'(r)=\frac{O}{2}-\frac{8}{3}\pi r^2=0\). Therefore, \(O=\frac{16}{3}\pi r^2\). 7. Substitute this relation into the expression for \(h\): \(h=\frac{\frac{16}{3}\pi r^2}{2\pi r}-\frac{4}{3}r=\frac{4}{3}r\). 8. Since \(V''(r)=-\frac{16}{3}\pi r<0\) for \(r>0\), the volume function is concave down and this critical point gives the maximum.

The volume is maximized when \(h=\frac{4}{3}r\). Thus the cylinder height equals the cone height, and the ratio is \(h:r=4:3\).

- Write the total volume and exterior area as sums for the cylinder and two cones. - Use the fixed volume to express one variable in terms of the other. - Use the Pythagorean theorem to find each cone's slant height. - Differentiate the one-variable surface-area function to find its minimum. - Substitute the critical-point relation back into the expression for \(h\) to obtain the ratio.

1. The total volume is \(V=\pi r^2h+2\left(\frac{1}{3}\pi r^3\right)=\pi r^2h+\frac{2}{3}\pi r^3\). 2. Each cone has slant height \(s=\sqrt{r^2+r^2}=r\sqrt{2}\). Therefore, the exterior surface area is \(A=2\pi rh+2\pi rs=2\pi rh+2\sqrt{2}\pi r^2\). 3. Solve the volume equation for \(h\): \(h=\frac{V}{\pi r^2}-\frac{2}{3}r\). 4. Substitute into the area function: \(A(r)=\frac{2V}{r}+\pi r^2\left(2\sqrt{2}-\frac{4}{3}\right)\). 5. Differentiate and set the derivative equal to zero: \(A'(r)=-\frac{2V}{r^2}+2\pi r\left(2\sqrt{2}-\frac{4}{3}\right)=0\). Thus \(V=\pi r^3\left(2\sqrt{2}-\frac{4}{3}\right)\). 6. Substitute this relation into the expression for \(h\): \(h=r\left(2\sqrt{2}-\frac{4}{3}\right)-\frac{2}{3}r=2(\sqrt{2}-1)r\). 7. The second derivative is positive for \(r>0\), so this critical point gives the minimum surface area.

The optimal ratio is \(h:r=2(\sqrt{2}-1):1\). Equivalently, \(h=2(\sqrt{2}-1)r\approx 0.828r\).

- Separate the greenhouse into a square prism and a square pyramid. - Include the floor, four walls, and four roof faces in \(S\). - Use the Pythagorean theorem to find the roof's slant height. - Solve the area constraint for \(h\), then maximize the resulting one-variable volume function.

1. The roof slant height is \(s=\sqrt{\left(\frac{a}{2}\right)^2+\left(\frac{a}{2}\right)^2}=\frac{a}{\sqrt{2}}\). Therefore, the four triangular roof faces have total area \(4\left(\frac{1}{2}as\right)=\sqrt{2}a^2\). 2. The fixed exterior surface area is \(S=a^2+4ah+\sqrt{2}a^2=a^2(1+\sqrt{2})+4ah\). The volume is \(V=a^2h+\frac{1}{3}a^2\left(\frac{a}{2}\right)=a^2h+\frac{1}{6}a^3\). 3. Solve the surface-area equation for \(h\): \(h=\frac{S}{4a}-\frac{1+\sqrt{2}}{4}a\). 4. Substitute into the volume: \(V(a)=\frac{S}{4}a-\frac{1+3\sqrt{2}}{12}a^3\). 5. Differentiate and set the derivative equal to zero: \(V'(a)=\frac{S}{4}-\frac{1+3\sqrt{2}}{4}a^2=0\). Thus \(a=\sqrt{\frac{S}{1+3\sqrt{2}}}\). 6. The second derivative is negative for \(a>0\), so this critical point gives the maximum. 7. Substitute \(S=(1+3\sqrt{2})a^2\) into the expression for \(h\): \(h=\frac{\sqrt{2}}{2}a\).

The maximizing dimensions are \(a=\sqrt{\frac{S}{1+3\sqrt{2}}}\) and \(h=\frac{\sqrt{2}}{2}a\).

- Use the Pythagorean theorem in an axial cross section. - Recall that a cylinder's lateral area does not include its bases. - Because the area is nonnegative, maximizing its square gives the same maximizing height. - Compare your maximum with the curved area of a hemisphere.

1. In an axial cross section, the cylinder radius \(r\), height \(h\), and hemisphere radius \(R\) form a right triangle: \(r^2+h^2=R^2\). Thus \(r=\sqrt{R^2-h^2}\), and the lateral area is \(M(h)=2\pi h\sqrt{R^2-h^2}\). 2. Because \(M(h)\ge 0\), maximize \(f(h)=\left(\frac{M}{2\pi}\right)^2=h^2(R^2-h^2)=R^2h^2-h^4\). Its derivative is \(f'(h)=2R^2h-4h^3=2h(R^2-2h^2)\). The interior critical point is \(h=\frac{R}{\sqrt{2}}\). Since \(f=0\) at \(h=0\) and \(h=R\), this critical point gives the maximum. 3. At this height, \(r=\sqrt{R^2-\frac{R^2}{2}}=\frac{R}{\sqrt{2}}\). Therefore, \(M_{\max}=2\pi\left(\frac{R}{\sqrt{2}}\right)\left(\frac{R}{\sqrt{2}}\right)=\pi R^2\). The hemisphere's curved surface area is \(2\pi R^2\), so \(\frac{M_{\max}}{2\pi R^2}=\frac{1}{2}\).

1. \(M(h)=2\pi h\sqrt{R^2-h^2}\) 2. The lateral area is maximized when \(h=\frac{R}{\sqrt{2}}\). 3. The maximum is \(M_{\max}=\pi R^2\), exactly one-half of the hemisphere's curved surface area \(2\pi R^2\).

- Draw the slab in a coordinate plane and write the break line's equation. - Let one corner of the new rectangle move along the break line. - Determine the allowed interval for that corner. - Check whether the critical point lies inside the interval; otherwise compare the endpoint behavior.

1. Place the original slab in the coordinate plane with corners \((0, 0)\), \((b, 0)\), \((b, a)\), and \((0, a)\), and let the broken corner be \((0, a)\). The break runs from \((0, \frac{3a}{4})\) to \((\frac{a}{4}, a)\), so its equation is \(y=x+\frac{3a}{4}\), where \(0\le x\le\frac{a}{4}\). 2. If the new rectangle's upper-left corner is \((x, y)\) on the break, its width and height are \(w=b-x\) and \(h=y=x+\frac{3a}{4}\). Thus \(A(x)=(b-x)\left(x+\frac{3a}{4}\right)\). 3. Differentiate: \(A'(x)=-2x+b-\frac{3a}{4}\). The unconstrained critical point is \(x_0=\frac{b}{2}-\frac{3a}{8}\). Since \(b\ge a\), this value is positive. It lies in the allowed interval when \(\frac{b}{2}-\frac{3a}{8}\le\frac{a}{4}\), which is equivalent to \(b\le\frac{5a}{4}\). 4. If \(a\le b\le\frac{5a}{4}\), the concave quadratic is maximized at \(x_0\). Then \(w=h=\frac{b}{2}+\frac{3a}{8}\). 5. If \(b>\frac{5a}{4}\), the critical point lies beyond the allowed interval, and \(A\) is increasing throughout that interval. The maximum occurs at \(x=\frac{a}{4}\), giving \(w=b-\frac{a}{4}\) and \(h=a\).

If \(a\le b\le\frac{5a}{4}\), the maximum-area piece is a square with side length \(\frac{b}{2}+\frac{3a}{8}\). If \(b>\frac{5a}{4}\), the maximum-area rectangle has dimensions \(b-\frac{a}{4}\) by \(a\).

- Write the prism volume using its square base edge and height. - Use similar triangles to find the cone's radius at the height of the prism's upper face. - Relate that radius to half the square's diagonal. - Check the critical values and endpoints.

1. The prism volume is \(V=a^2h\). At height \(h\), the cone's cross-sectional radius is related by similar triangles. The distance from the center of the square to a vertex is \(\frac{a\sqrt{2}}{2}\), so \(\frac{a\sqrt{2}/2}{12}=\frac{18-h}{18}\). Therefore, \(a=\frac{2\sqrt{2}}{3}(18-h)\). 2. Substitute into the volume: \(V(h)=\frac{8}{9}(18-h)^2h\), where \(0\le h\le18\). 3. Differentiate: \(V'(h)=\frac{8}{3}(h^2-24h+108)\). The critical values are \(h=6\) and \(h=18\). The value \(h=18\) is an endpoint with zero volume, while \(h=6\,\text{cm}\) gives the interior maximum. 4. The base edge length is \(a=\frac{2\sqrt{2}}{3}(18-6)=8\sqrt{2}\,\text{cm}\approx 11.31\,\text{cm}\). 5. The maximum volume is \(V=(8\sqrt{2})^2(6)=768\,\text{cm}^3\).

The maximum-volume prism has base edge length \(a=8\sqrt{2}\,\text{cm}\approx 11.31\,\text{cm}\) and height \(h=6\,\text{cm}\). Its maximum volume is \(768\,\text{cm}^3\).

- Use a right triangle whose hypotenuse is the distance \(s\). - Use the cylinder volume to eliminate one variable. - Minimize \(s^2\) instead of \(s\). - Rewrite the critical-point equation using \(V=\pi r^2h\).

1. The distance satisfies \(s^2=r^2+h^2\). Because \(s>0\), minimizing \(s\) is equivalent to minimizing \(s^2\). 2. The fixed-volume condition is \(V=\pi r^2h\), so \(h=\frac{V}{\pi r^2}\). 3. Substitute into the squared distance: \(f(r)=r^2+\frac{V^2}{\pi^2r^4}\). 4. Differentiate and set the derivative equal to zero: \(f'(r)=2r-\frac{4V^2}{\pi^2r^5}=0\). Thus \(r^6=\frac{2V^2}{\pi^2}\). 5. Since \(V=\pi r^2h\), substitute \(V^2=\pi^2r^4h^2\): \(r^6=2r^4h^2\). Therefore, \(r^2=2h^2\), so \(\frac{r}{h}=\sqrt{2}\). The objective approaches infinity at both ends of its positive domain, so the critical point gives the absolute minimum.

The optimal ratio is \(r:h=\sqrt{2}:1\), or \(r=\sqrt{2}h\).

- Use the Pythagorean theorem to relate \(r\), \(h\), and \(s\). - Use the fixed volume to eliminate one variable. - Minimize \(s^2\) rather than \(s\). - Express the critical-point equation as a ratio of \(h\) and \(r\).

1. The slant height satisfies \(s^2=r^2+h^2\). Since \(s>0\), minimize \(s^2\). 2. The fixed-volume condition is \(V=\frac{1}{3}\pi r^2h\), so \(r^2=\frac{3V}{\pi h}\). 3. Substitute into the squared slant height: \(f(h)=\frac{3V}{\pi h}+h^2\). 4. Differentiate and set the derivative equal to zero: \(f'(h)=-\frac{3V}{\pi h^2}+2h=0\). Therefore, \(2h^3=\frac{3V}{\pi}\). 5. Using \(V=\frac{1}{3}\pi r^2h\), this becomes \(2h^3=r^2h\). Hence \(r^2=2h^2\), and \(h:r=1:\sqrt{2}\). The objective approaches infinity as \(h\to0^+\) or \(h\to\infty\), so the critical point gives the absolute minimum.

The slant height is minimized when \(h:r=1:\sqrt{2}\), or \(r=\sqrt{2}h\).

- Write the pyramid volume and total surface-area formulas. - Express each triangular face's slant height using \(a\) and \(h\). - Squaring the nonnegative volume can simplify the optimization. - Determine the physically possible interval for the base edge.

1. The volume is \(V=\frac{1}{3}a^2h\). The total surface area is \(a^2+a\sqrt{4h^2+a^2}=432\). 2. Isolate the radical and square: \((432-a^2)^2=a^2(4h^2+a^2)\). This gives \(h^2=\frac{432^2}{4a^2}-216\). 3. Because \(V\ge0\), maximize \(V^2\). Let \(u=a^2\). Then \(V^2=\frac{1}{9}a^4h^2=5184u-24u^2\). 4. Differentiate: \(\frac{d}{du}(V^2)=5184-48u\). The critical point is \(u=108\). Since the quadratic has negative leading coefficient, this critical point gives the absolute maximum on the physical domain. 5. Therefore, \(a=\sqrt{108}=6\sqrt{3}\,\text{cm}\approx 10.39\,\text{cm}\). Also, \(h^2=\frac{432^2}{4(108)}-216=216\), so \(h=6\sqrt{6}\,\text{cm}\approx 14.70\,\text{cm}\).

The volume is maximized when \(a=6\sqrt{3}\,\text{cm}\approx 10.39\,\text{cm}\) and \(h=6\sqrt{6}\,\text{cm}\approx 14.70\,\text{cm}\).

- Use similar triangles to express each inner solid's height in terms of its horizontal dimension. - For the square prism, relate the square's side length to the radius of its circumscribed circle. - Differentiate each one-variable volume function. - In the final ratio, \(R^2H\) cancels.

1. For a cylinder of radius \(r\) and height \(h\), similar triangles give \(\frac{r}{R}=\frac{H-h}{H}\), so \(h=H\left(1-\frac{r}{R}\right)\). The volume is \(V(r)=\pi Hr^2\left(1-\frac{r}{R}\right)\). Differentiate: \(V'(r)=\pi H\left(2r-\frac{3r^2}{R}\right)\). The interior critical point is \(r=\frac{2R}{3}\), which gives \(h=H\left(1-\frac{2}{3}\right)=\frac{H}{3}\). The endpoint volumes are zero, so this critical point gives the maximum. 2. Let \(s\) be the square base edge. The distance from the square's center to a vertex is \(\frac{s\sqrt{2}}{2}\). Similar triangles give \(h=H\left(1-\frac{s\sqrt{2}}{2R}\right)\). Thus \(V(s)=Hs^2\left(1-\frac{s\sqrt{2}}{2R}\right)\). Differentiating gives the interior critical point \(s=\frac{2\sqrt{2}}{3}R\). At this value, \(h=\frac{H}{3}\), so \(V_{\text{prism, max}}=\left(\frac{2\sqrt{2}}{3}R\right)^2\left(\frac{H}{3}\right)=\frac{8}{27}R^2H\). 3. The maximum cylinder volume is \(V_{\text{cyl, max}}=\pi\left(\frac{2R}{3}\right)^2\left(\frac{H}{3}\right)=\frac{4\pi}{27}R^2H\). Therefore, \(\frac{V_{\text{cyl, max}}}{V_{\text{prism, max}}}=\frac{\pi}{2}\approx1.57\). The cylinder has the larger maximum volume.

1. The maximum-volume cylinder has height \(h=\frac{H}{3}\). 2. The maximum square-prism volume is \(V_{\text{prism, max}}=\frac{8}{27}R^2H\). 3. The ratio is \(\frac{V_{\text{cyl, max}}}{V_{\text{prism, max}}}=\frac{\pi}{2}\approx1.57\), so the cylinder uses the available space more efficiently.

- Write the cone volume and lateral-area formulas. - Relate \(r\), \(h\), and the slant height \(s\). - Use the fixed lateral area to eliminate \(s\). - Maximize the expression under the square root.

1. The volume is \(V=\frac{1}{3}\pi r^2h\). The fixed lateral area is \(M=\pi rs\), where \(s=\sqrt{r^2+h^2}\). 2. From the area constraint, \(s=\frac{M}{\pi r}\). Therefore, \(h=\sqrt{\frac{M^2}{\pi^2r^2}-r^2}\). 3. Substitute into the volume: \(V(r)=\frac{1}{3}\sqrt{M^2r^2-\pi^2r^6}\). Because the square root is increasing, maximize \(f(r)=M^2r^2-\pi^2r^6\). 4. Differentiate: \(f'(r)=2M^2r-6\pi^2r^5\). At the positive critical point, \(M^2=3\pi^2r^4\), so \(M=\pi r^2\sqrt{3}\). 5. Since \(M=\pi rs\), it follows that \(s=r\sqrt{3}\). Thus \(h=\sqrt{s^2-r^2}=r\sqrt{2}\). The objective is zero at the endpoints of its physical domain, so this critical point gives the maximum.

The volume is maximized when \(\frac{h}{r}=\sqrt{2}\), or \(h:r=\sqrt{2}:1\).

- Write a squared-distance function for \((x, f(x))\) and \(S\). - Use the parabola's symmetry to check your results. - Find and classify every critical point. - Compare with the distance from the vertex.

1. A point on the graph is \((x, 0.5x^2-1.5)\). Its squared distance from \(S\) is \(D(x)=x^2+(0.5x^2-2.5)^2=0.25x^4-1.5x^2+6.25\). 2. Differentiate: \(D'(x)=x^3-3x=x(x^2-3)\). The critical values are \(x=0\) and \(x=\pm\sqrt{3}\). 3. The second derivative is \(D''(x)=3x^2-3\). Thus \(x=0\) gives a local maximum, while \(x=\pm\sqrt{3}\) give local minima. Since \(D(x)\to\infty\) as \(|x|\to\infty\), these are the absolute minima. 4. At \(x=\pm\sqrt{3}\), \(f(x)=0\). The minimum distance is \(d=\sqrt{(\sqrt{3})^2+(0-1)^2}=2\).

The closest points are \(P_1(-\sqrt{3}, 0)\) and \(P_2(\sqrt{3}, 0)\). The minimum distance is \(2\) units.

- Simplify the objective with a double-angle identity before differentiating. - Ignore positive constant factors when locating the maximum. - Use the product and chain rules. - Check the endpoint behavior.

1. Multiply the two expressions: \(P(\alpha)=\frac{v^4}{2g^2}\sin(2\alpha)\sin^2(\alpha)\). Using \(\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)\), \(P(\alpha)=\frac{v^4}{g^2}\sin^3(\alpha)\cos(\alpha)\). 2. The constant factor does not affect the maximizing angle, so maximize \(f(\alpha)=\sin^3(\alpha)\cos(\alpha)\). 3. Differentiate: \(f'(\alpha)=3\sin^2(\alpha)\cos^2(\alpha)-\sin^4(\alpha)\) \(=\sin^2(\alpha)\left(3\cos^2(\alpha)-\sin^2(\alpha)\right)\). 4. The interior critical point satisfies \(3\cos^2(\alpha)=\sin^2(\alpha)\), so \(\tan^2(\alpha)=3\). In the given interval, \(\alpha=60^\circ\). 5. The product is zero at \(0^\circ\) and approaches zero as \(\alpha\to90^\circ-\), so \(60^\circ\) gives the absolute maximum.

The product of range and maximum height is maximized at \(\alpha=60^\circ\).

- Rewrite \(\frac{k}{x}\) using a negative exponent before differentiating. - At a critical point, use the derivative equation to replace \(k\). - Use the distance formula and square both sides. - A substitution for \(x^2\) turns the resulting equation into a quadratic.

1. Differentiate: \(f_k'(x)=2x-\frac{k}{x^2}\) and \(f_k''(x)=2+\frac{2k}{x^3}>0\). The critical-point equation gives \(2x^3=k\), so \(x=\sqrt[3]{\frac{k}{2}}\). Thus the critical point is a local minimum. 2. At this point, \(k=2x^3\), so \(y=x^2+\frac{2x^3}{x}=3x^2\). Therefore, the local minimum is \(\left(\sqrt[3]{\frac{k}{2}}, 3\left(\frac{k}{2}\right)^{2/3}\right)\), and the locus is \(y=3x^2\) for \(x>0\). 3. On the locus, the squared distance from the origin is \(x^2+y^2=x^2+9x^4\). Setting this equal to \(10\) gives \(9x^4+x^2-10=0\). 4. Let \(u=x^2\). Then \(9u^2+u-10=(u-1)(9u+10)=0\). Since \(x>0\), \(u=1\) and \(x=1\). From \(k=2x^3\), \(k=2\).

a) \(\left(\sqrt[3]{\frac{k}{2}}, 3\left(\frac{k}{2}\right)^{2/3}\right)\) b) \(y=3x^2\) for \(x>0\) c) \(k=2\)

- Use time equals distance divided by speed. - Express the forest and road distances using one variable. - Differentiate the total-time function. - Convert the fractional part of an hour to minutes.

1. Let \(y\) be the distance from \(C\) to \(P\). The forest distance is \(\sqrt{4^2+y^2}\), and the road distance is \(10-y\). 2. The travel-time function is \(T(y)=\frac{\sqrt{16+y^2}}{3}+\frac{10-y}{5}\), where \(0\le y\le10\). 3. Differentiate: \(T'(y)=\frac{y}{3\sqrt{16+y^2}}-\frac{1}{5}\). Setting the derivative equal to zero gives \(5y=3\sqrt{16+y^2}\). Squaring yields \(25y^2=9(16+y^2)\), so \(y=3\). 4. Compare the critical point and endpoints: \(T(0)=\frac{10}{3}\), \(T(10)=\frac{\sqrt{116}}{3}\), and \(T(3)=\frac{5}{3}+\frac{7}{5}=\frac{46}{15}\) hours. Therefore, the minimum occurs at \(y=3\) miles. 5. Since \(\frac{46}{15}=3+\frac{1}{15}\), the travel time is \(3\) hours and \(4\) minutes.

The hiker should reach the road \(3\) miles from \(C\) toward \(T\). The minimum travel time is \(3\) hours \(4\) minutes.

- Use the relationship between a circle's radius and circumference. - Express the volume as a function of the height. - Keep the height restriction in the domain. - Compare the interior critical point with both endpoints.

1. The volume is \(V=\pi r^2h\). The constraint is \(h+2\pi r=180\), so \(r=\frac{180-h}{2\pi}\). 2. Substitute into the volume: \(V(h)=\frac{(180-h)^2h}{4\pi}\), where \(20\le h\le100\). 3. Differentiate: \(V'(h)=\frac{3h^2-720h+32{,}400}{4\pi}\). The critical values are \(h=60\) and \(h=180\). Only \(h=60\) lies in the allowed interval. 4. Compare the critical point and endpoints: \(V(20)=\frac{128{,}000}{\pi}\approx40{,}743.67\,\text{cm}^3\), \(V(60)=\frac{216{,}000}{\pi}\approx68{,}754.94\,\text{cm}^3\), and \(V(100)=\frac{160{,}000}{\pi}\approx50{,}929.58\,\text{cm}^3\). Thus the absolute maximum occurs at \(h=60\,\text{cm}\). 5. The radius is \(r=\frac{180-60}{2\pi}=\frac{60}{\pi}\,\text{cm}\approx19.10\,\text{cm}\).

a) The maximizing dimensions are \(h=60\,\text{cm}\) and \(r=\frac{60}{\pi}\,\text{cm}\approx19.10\,\text{cm}\). b) The maximum volume is \(\frac{216{,}000}{\pi}\,\text{cm}^3\approx68{,}754.94\,\text{cm}^3\).

- Use the three zeros to determine where the function is positive and negative. - Add the absolute values of the two definite integrals. - Simplify the total area as a function of \(a\). - Differentiate the area function, and use convexity or the second derivative to verify the minimum.

1. The zeros are \(x=0\), \(x=a\), and \(x=1\). The function is positive on \((0, a)\) and negative on \((a, 1)\). 2. An antiderivative is \(F_a(x)=\frac14x^4-\frac{a+1}{3}x^3+\frac{a}{2}x^2\). Therefore, the total area is \(A(a)=\int_0^a f_a(x)\, \mathrm{d}x-\int_a^1 f_a(x)\, \mathrm{d}x\) \(=2F_a(a)-F_a(1)\) \(=\frac{-2a^4+4a^3-2a+1}{12}\). 3. Differentiate: \(A'(a)=\frac{-4a^3+6a^2-1}{6}\) \(=\frac{(2a-1)(-2a^2+2a+1)}{6}\). The only critical point in \((0, 1)\) is \(a=\frac12\). 4. The second derivative is \(A''(a)=2a(1-a)>0\) throughout \((0, 1)\), so \(A\) is strictly convex and \(a=\frac12\) gives the absolute minimum. 5. The minimum total area is \(A\left(\frac12\right)=\frac1{32}\).

The total area is minimized when \(a=\frac12\). The minimum area is \(\frac1{32}\) square unit.

- Find the zero of \(f_t\) in terms of \(t\). - First analyze the case in which the zero lies inside \([0, 1]\). - Split the area integral at the zero and simplify. - Check the parameter ranges in which the zero lies outside the interval.

1. The zero of \(f_t\) is \(x=\ln t\). This zero lies in \([0, 1]\) when \(1\le t\le e\). 2. For \(1\le t\le e\), the total area is \(A(t)=\int_0^{\ln t}(t-e^x)\, \mathrm{d}x+\int_{\ln t}^1(e^x-t)\, \mathrm{d}x\) \(=2t\ln t-3t+e+1\). 3. Differentiate: \(A'(t)=2\ln t-1\). The critical point is \(t=\sqrt e\). Since \(A''(t)=\frac2t>0\), this is a local minimum in \([1, e]\). 4. For \(0<t\le1\), the graph stays above the x-axis, so \(A(t)=e-1-t\), whose smallest value on that range is \(A(1)=e-2\). For \(t\ge e\), the graph stays below the x-axis, so \(A(t)=t-e+1\), whose smallest value on that range is \(A(e)=1\). 5. At the interior critical point, \(A(\sqrt e)=e+1-2\sqrt e\approx0.421\). This is smaller than both boundary-range minima, so it is the absolute minimum for \(t>0\).

The total area is minimized when \(t=\sqrt e\). The minimum area is \(e+1-2\sqrt e\approx0.421\) square unit.

- Set the two functions equal to find their intersections. - Factor the difference between the functions. - Integrate the upper function minus the lower function between the intersections. - Differentiate the area function with respect to \(k\). - Verify that the critical point gives the absolute maximum.

1. Set the functions equal: \(4k^2-4x^2=k^3-kx^2\). Rearranging gives \((4-k)(k^2-x^2)=0\). Since \(0<k<4\), it follows that \(x^2=k^2\), so the intersections occur at \(x=\pm k\). 2. On \([-k, k]\), \(f_k(x)-g_k(x)=(4-k)(k^2-x^2)\ge0\). Therefore, \(A(k)=\int_{-k}^k(4-k)(k^2-x^2)\, \mathrm{d}x\) \(=\frac43(4k^3-k^4)\) \(=\frac{16}{3}k^3-\frac43k^4\). 3. Differentiate: \(A'(k)=16k^2-\frac{16}{3}k^3\) \(=16k^2\left(1-\frac{k}{3}\right)\). The only critical point in \((0, 4)\) is \(k=3\). Since \(A''(3)=32(3)-16(3)^2=-48<0\), this point is a local maximum. Also, \(A(k)\to0\) at both ends of the domain, so it is the absolute maximum. 4. The maximum area is \(A(3)=\frac43(108-81)=36\) square units.

a) The graphs intersect at \(x=\pm k\). b) \(A(k)=\frac{16}{3}k^3-\frac43k^4\) c) The area is maximized at \(k=3\), and the maximum area is \(36\) square units.

- Set the constant function equal to the quadratic function. - Use symmetry when setting up the area integral. - Integrate the upper graph minus the lower graph. - Differentiate the area function with respect to \(a\). - Substitute the maximizing value into the area formula.

1. Set the functions equal: \(a^2=\frac{a^2}{(6-a)^2}x^2\). Since \(a>0\), divide by \(a^2\) to obtain \(x^2=(6-a)^2\). Thus, the intersections are \((-(6-a), a^2)\) and \((6-a, a^2)\). 2. The horizontal line \(f_a\) lies above the parabola between the intersections, so \(A(a)=\int_{-(6-a)}^{6-a}\left(a^2-\frac{a^2}{(6-a)^2}x^2\right)\, \mathrm{d}x\) \(=\frac43a^2(6-a)\) \(=8a^2-\frac43a^3\). 3. Differentiate: \(A'(a)=16a-4a^2=4a(4-a)\). The critical point in \((0, 6)\) is \(a=4\). Since \(A''(4)=16-8(4)=-16<0\), this is a local maximum. The area approaches \(0\) at both ends of the domain, so it is also the absolute maximum. 4. The maximum area is \(A(4)=\frac43(4^2)(2)=\frac{128}{3}\) square units.

a) The intersection points are \((-(6-a), a^2)\) and \((6-a, a^2)\). b) \(A(a)=\frac43a^2(6-a)=8a^2-\frac43a^3\) c) The area is maximized at \(a=4\), and the maximum area is \(\frac{128}{3}\) square units.

- Find the intersection points by setting the equations equal. - Determine which parabola is above the other between the intersections. - Integrate the difference between the functions. - Differentiate the resulting area function with respect to \(k\).

1. Set the equations equal: \(kx^2-4k=\frac4{k^2}-\frac1{k^2}x^2\). This becomes \(\left(k+\frac1{k^2}\right)x^2=4\left(k+\frac1{k^2}\right)\). Since the factor in parentheses is positive, \(x=\pm2\). Both intersection points lie on the x-axis. 2. On \([-2, 2]\), \(q_k\) lies above \(p_k\). Thus, \(A(k)=\int_{-2}^2\left[\left(\frac4{k^2}+4k\right)-\left(\frac1{k^2}+k\right)x^2\right]\, \mathrm{d}x\) \(=\frac{32}{3}\left(k+\frac1{k^2}\right)\). 3. Differentiate: \(A'(k)=\frac{32}{3}\left(1-\frac2{k^3}\right)\). The critical point is \(k=\sqrt[3]{2}\). Since \(A''(k)=\frac{64}{k^4}>0\), this is a local minimum. Also, \(A(k)\to\infty\) as \(k\to0^+\) and as \(k\to\infty\), so it is the absolute minimum. 4. The minimum area is \(A(\sqrt[3]{2})=16\sqrt[3]{2}\approx20.16\) square units.

The area is minimized when \(k=\sqrt[3]{2}\). The minimum area is \(16\sqrt[3]{2}\approx20.16\) square units.

- Find the two intersection points in terms of \(a\). - Integrate the line minus the parabola between the intersections. - Simplify the area as a function of \(a\). - Differentiate and analyze the sign of the derivative.

1. Set the functions equal: \(ax^2=(a+1)x\). The intersection x-coordinates are \(x=0\) and \(x=\frac{a+1}{a}\). 2. The line lies above the parabola between the intersections, so \(A(a)=\int_0^{(a+1)/a}\left[(a+1)x-ax^2\right]\, \mathrm{d}x\) \(=\frac{(a+1)^3}{6a^2}\). 3. Differentiate: \(A'(a)=\frac{(a+1)^2(a-2)}{6a^3}\). The only positive critical point is \(a=2\). The derivative is negative for \(0<a<2\) and positive for \(a>2\), so this point gives the absolute minimum. 4. The minimum area is \(A(2)=\frac{27}{24}=\frac98\) square units.

The area is minimized when \(a=2\). The minimum area is \(\frac98=1.125\) square units.

- Find the line's slope from the two given points. - Subtract the line from the logarithmic function. - Differentiate using \(\frac{d}{dx}\ln(x+1)=\frac{1}{x+1}\). - Keep both the maximizing x-value and maximum difference exact.

1. The slope of the line through \((0, 0)\) and \((e-1, 1)\) is \(m=\frac{1}{e-1}\). Therefore, \(g(x)=\frac{x}{e-1}\). 2. The difference function is \(d(x)=\ln(x+1)-\frac{x}{e-1}\). 3. Differentiate: \(d'(x)=\frac{1}{x+1}-\frac{1}{e-1}\). Setting the derivative equal to zero gives \(x=e-2\). 4. Since \(d''(x)=-\frac{1}{(x+1)^2}<0\), this critical point gives the maximum. The endpoint values are both zero. 5. The exact maximum difference is \(d(e-2)=\ln(e-1)-\frac{e-2}{e-1}\).

a) \(g(x)=\frac{x}{e-1}\) b) The difference is greatest at \(x=e-2\). Its maximum value is \(\ln(e-1)-\frac{e-2}{e-1}\).

- Write the squared distance from \((x, f(x))\) to the origin. - Differentiate using the chain rule and the derivative of \(\ln x\). - Study the monotonicity of the expression inside the factored derivative. - Check the behavior as \(x\to0^+\) and \(x\to\infty\).

1. A point on the graph is \((x, 1-\ln x)\). Minimize the squared distance \(g(x)=x^2+(1-\ln x)^2\). 2. Differentiate: \(g'(x)=2x-\frac{2}{x}+\frac{2\ln x}{x}=\frac{2}{x}(x^2-1+\ln x)\). 3. Let \(q(x)=x^2-1+\ln x\). Then \(q'(x)=2x+\frac{1}{x}>0\) for \(x>0\), so \(q\) is strictly increasing. Since \(q(1)=0\), the only critical point is \(x=1\). 4. Also, \(g(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\). Therefore, \(x=1\) gives the absolute minimum. 5. The corresponding y-coordinate is \(f(1)=1\).

The closest point is \(P(1, 1)\).

- Write the tangent line at \(x=a\). - Find each axis intercept by setting one coordinate equal to \(0\). - Use an absolute value when writing the rectangle's area. - For a maximum, find critical points and check the behavior at the ends of the interval.

1. Since \(g'(x)=\frac{2}{x}\), the tangent line at \(x=a\) is \(y=\frac{2}{a}(x-a)+2\ln(a)=\frac{2}{a}x-2+2\ln(a)\). 2. Set \(x=0\) to find the y-intercept: \(R=(0, 2\ln(a)-2)\). 3. Set \(y=0\) to find the x-intercept: \(0=\frac{2}{a}x-2+2\ln(a)\), so \(S=(a(1-\ln(a)), 0)\). 4. For \(0<a<e\), \(x_S>0\) and \(y_R<0\). Therefore, \(A(a)=|x_Sy_R|=2a(1-\ln(a))^2\). 5. Differentiate: \(A'(a)=2(1-\ln(a))^2-4(1-\ln(a))=2(1-\ln(a))(-1-\ln(a))\). 6. The critical values satisfy \(1-\ln(a)=0\) or \(-1-\ln(a)=0\). The first gives \(a=e\), which is not in the interval. The interior critical point is \(a=\frac{1}{e}\). 7. Also, \(A(a)\to0\) as \(a\to0^+\) and as \(a\to e^-\). Thus the interior critical point gives the global maximum. Its value is \(A\left(\frac{1}{e}\right)=\frac{8}{e}\approx2.943\) square units.

a) \(S=(a(1-\ln(a)), 0)\); \(R=(0, 2\ln(a)-2)\) b) \(A(a)=2a(1-\ln(a))^2\) c) The maximum area is \(\frac{8}{e}\approx2.943\) square units at \(a=\frac{1}{e}\).

- Find the second derivative of \(f\) first. - A vertical distance is the absolute difference of the two function values. - Check the sign of the difference on the stated domain before removing the absolute value. - Maximize the resulting distance function with a derivative test.

1. Differentiate twice: \(f'(x)=(-x^2+4x+1)e^{-x}\) and \(f''(x)=(x^2-6x+3)e^{-x}\). 2. The vertical distance is \(L(a)=|f(a)-f''(a)|\). The difference simplifies to \((4a-6)e^{-a}\). Since \(a>\frac{3}{2}\), this expression is positive, so \(L(a)=(4a-6)e^{-a}\). 3. Differentiate: \(L'(a)=(10-4a)e^{-a}\). The only critical number in the domain is \(a=\frac{5}{2}\). 4. The derivative is positive for \(\frac{3}{2}<a<\frac{5}{2}\) and negative for \(a>\frac{5}{2}\). Therefore, \(L\) reaches its absolute maximum at \(a=\frac{5}{2}\).

\(a=\frac{5}{2}\).

- Express the glide time in seconds and use it to find the altitude lost. - Divide the lost altitude by the climb rate to find the climbing time. - Remove positive constant factors before minimizing. - Use the first derivative to find the optimal speed. - Write the tangent line to \(g\) at the speed found in part b.

1. The glide time is \(\frac{s}{c}\) hours, or \(t_{\mathrm{glide}}=\frac{3600s}{c}\) seconds. The altitude lost is \(g(c)t_{\mathrm{glide}}\), so the climbing time is \(t_{\mathrm{climb}}=\frac{g(c)}{v_{\mathrm{climb}}}\cdot\frac{3600s}{c}\). Therefore, \(T(c)=\frac{3600s}{c}\left(1+\frac{g(c)}{v_{\mathrm{climb}}}\right)\). 2. Since the positive factors \(3600s\) and \(\frac{1}{v_{\mathrm{climb}}}\) do not affect the minimizing value, minimize \(F(c)=\frac{v_{\mathrm{climb}}+g(c)}{c}\). 3. Substitute the model: \(F(c)=\frac{3.2+0.0005(c-80)^2+0.8}{c}=0.0005c-0.08+\frac{7.2}{c}\). 4. Differentiate: \(F'(c)=0.0005-\frac{7.2}{c^2}\). Setting \(F'(c)=0\) gives \(c^2=14{,}400\). Since \(c>0\), \(c=120\,\text{km/h}\). Also, \(F''(120)=\frac{14.4}{120^3}>0\), so this is a minimum. 5. At \(c=120\), \(g(120)=1.6\) and \(g'(120)=0.001(120-80)=0.04\). The tangent line is \(y=0.04(c-120)+1.6=0.04c-3.2\). 6. At \(c=0\), this tangent line has value \(-3.2=-v_{\mathrm{climb}}\), so it passes through \(P(0, -v_{\mathrm{climb}})\) and touches the graph at the optimal speed.

a) \(T(c)=\frac{3600s}{c}\left(1+\frac{g(c)}{v_{\mathrm{climb}}}\right)\) b) \(c=120\,\text{km/h}\) c) The tangent line at \((120, 1.6)\) is \(y=0.04c-3.2\), so it passes through \((0, -3.2)\).

- Use the two-point slope formula to relate the intercepts. - Apply the Pythagorean theorem to the intercept triangle. - Simplify the derivative before solving the critical-point equation. - Use the monotonicity of \(z^2\) for \(z>0\).

1. The line through \((x_0, 0)\) and \((1, 8)\) has slope \(m=\frac{8}{1-x_0}\). Its y-intercept is \(y_0=\frac{8x_0}{x_0-1}\). Therefore, by the Pythagorean theorem, \(L^2=x_0^2+y_0^2=x_0^2+\frac{64x_0^2}{(x_0-1)^2}\). 2. Let \(f(x_0)=L^2(x_0)\). Its derivative simplifies to \(f'(x_0)=2x_0\left(1-\frac{64}{(x_0-1)^3}\right)\). Since \(x_0>1\), the critical point satisfies \((x_0-1)^3=64\), so \(x_0=5\). 3. Also, \(f(x_0)\to\infty\) as \(x_0\to1^+\) or \(x_0\to\infty\), and this is the only critical point. Thus it gives the absolute minimum. 4. Because \(L(x_0)>0\) and the squaring function is strictly increasing for positive inputs, \(L\) and \(L^2\) attain their minima at the same value of \(x_0\).

a) \(L^2(x_0)=x_0^2+\frac{64x_0^2}{(x_0-1)^2}\) b) The hypotenuse is minimized when \(x_0=5\). c) Since \(L>0\) and squaring is strictly increasing on positive numbers, minimizing \(L^2\) also minimizes \(L\).

- Write the total area as rectangle area plus semicircle area. - Use the area constraint to express one dimension in terms of the other. - Account for the different costs of straight and curved frame sections. - Verify the minimum using the second derivative and the feasible domain.

1. Let \(r=\frac{b}{2}\) be the semicircle's radius. The area constraint is \(2rh+\frac{1}{2}\pi r^2=27\). 2. Solve for height: \(h=\frac{13.5}{r}-\frac{\pi r}{4}\). 3. The straight frame length is \(2h+2r\), and the curved length is \(\pi r\). Thus the cost is \(C=12(2h+2r)+18\pi r\). 4. Substitute for \(h\): \(C(r)=\frac{324}{r}+24r+12\pi r\). 5. Differentiate: \(C'(r)=-\frac{324}{r^2}+24+12\pi\). Setting this equal to zero gives \(r^2=\frac{27}{2+\pi}\). 6. Therefore, \(r\approx 2.29\,\text{ft}\), so \(b=2r\approx 4.58\,\text{ft}\). 7. The rectangular height is \(h=\frac{13.5}{r}-\frac{\pi r}{4}\approx 4.09\,\text{ft}\). 8. Since \(C''(r)=\frac{648}{r^3}>0\), this critical point gives the minimum cost.

The cost is minimized when the rectangular part is approximately \(4.58\,\text{ft}\) wide and \(4.09\,\text{ft}\) high.

- Multiply \(x\) by \(f(x)\) to form the area function. - Determine the first-quadrant domain from the intercepts. - Use the quotient rule and solve the resulting quadratic equation. - Check the endpoints.

1. The area is \(A(x)=xf(x)=\frac{8x-2x^2}{x+2}\). The graph meets the x-axis at \(x=4\), so \(0\le x\le4\). 2. Differentiate: \(A'(x)=\frac{-2x^2-8x+16}{(x+2)^2}\). The numerator is zero when \(x^2+4x-8=0\). The only solution in the domain is \(x=2\sqrt{3}-2\approx1.46\). 3. The endpoint areas are zero, and the derivative changes from positive to negative at the interior critical point. Therefore, it gives the absolute maximum. 4. Substitution gives \(A_{\max}=16-8\sqrt{3}\approx2.14\).

The area is maximized at \(x=2\sqrt{3}-2\approx1.46\). The maximum area is \(16-8\sqrt{3}\approx2.14\) square units.

- Expand the equation of \(C_t\) and subtract the two circle equations. - Write the resulting line in standard form. - Use the point-to-line distance formula for the origin. - Differentiate the distance function and apply the second derivative test.

1. Write the circle equations as \(x^2+y^2=32\) and \((x-t)^2+(y-t)^2=16\). Subtracting the equations eliminates the quadratic terms and gives \(2tx+2ty-2t^2=16\). Since \(t>0\), the radical axis is \(x+y=t+\frac8t\). 2. In standard form, the line is \(x+y-t-\frac8t=0\). Its distance from the origin is \(d(t)=\frac{t+\frac8t}{\sqrt2}\). 3. Differentiate: \(d'(t)=\frac1{\sqrt2}\left(1-\frac8{t^2}\right)\). Setting the derivative equal to zero gives \(t=2\sqrt2\). 4. The second derivative is \(d''(t)=\frac{16}{\sqrt2\, t^3}\). Therefore, \(d''(2\sqrt2)=\frac12>0\), so the extremum is a local minimum. 5. The minimum distance is \(d(2\sqrt2)=4\) units.

a) \(x+y=t+\frac8t\) b) The distance has a local extremum at \(t=2\sqrt2\). c) The distance is \(4\) units. Since \(d''(2\sqrt2)=\frac12>0\), the extremum is a local minimum.

- Find the cone's slant height from its radius and height. - Use the fixed volume to eliminate \(h\). - Include the bottom, cylinder side, and cone side in the area. - Rewrite the critical-point equation as a ratio.

1. The total volume is \(V=\pi r^2h+\frac{1}{3}\pi r^3\). Thus \(h=\frac{V}{\pi r^2}-\frac{r}{3}\). 2. The exterior surface consists of the bottom, cylindrical side, and conical lateral surface. The cone slant height is \(r\sqrt{2}\), so \(A=(1+\sqrt{2})\pi r^2+2\pi rh\). 3. Substitute for \(h\): \(A(r)=\left(\frac{1}{3}+\sqrt{2}\right)\pi r^2+\frac{2V}{r}\). 4. Differentiate and set the derivative equal to zero: \(A'(r)=2\left(\frac{1}{3}+\sqrt{2}\right)\pi r-\frac{2V}{r^2}=0\). Therefore, \(V=\left(\frac{1}{3}+\sqrt{2}\right)\pi r^3\). 5. Substitute this relation into the height formula: \(h=\left(\frac{1}{3}+\sqrt{2}\right)r-\frac{r}{3}=\sqrt{2}r\). 6. The second derivative is positive for \(r>0\), and the area approaches infinity at both ends of the positive domain, so this critical point gives the absolute minimum.

The optimal ratio is \(\frac{h}{r}=\sqrt{2}\), or \(h:r=\sqrt{2}:1\).

- Minimize the square of the distance instead of the distance itself. - Use the condition for perpendicular nonvertical lines: their slopes multiply to \(-1\). - Find the slope of the segment through the origin and \(P\). - The derivative gives the tangent slope.

1. Minimize the square of the distance: \(D^2(x)=x^2+\left(\frac{9}{x}\right)^2=x^2+81x^{-2}\). 2. Differentiate: \((D^2)'(x)=2x-162x^{-3}\). Setting this equal to zero gives \(2x=\frac{162}{x^3}\), so \(x^4=81\). Since \(x>0\), \(x=3\). 3. The second derivative is \((D^2)''(x)=2+486x^{-4}>0\), so the critical point is a minimum. Since \(f(3)=3\), the point is \(P(3, 3)\). 4. The slope of \(\overline{OP}\) is \(\frac{3-0}{3-0}=1\). 5. Since \(f'(x)=-\frac{9}{x^2}\), the tangent slope at \(x=3\) is \(-1\). 6. The product of the slopes is \(1\cdot(-1)=-1\), so the segment and tangent line are perpendicular.

1. \(P(3, 3)\) 2. The slope of \(\overline{OP}\) is \(1\), and the tangent slope is \(-1\); therefore, they are perpendicular.

- Minimize the squared distance from a general graph point to \(P\). - Differentiate and clear the denominators. - Test small positive integer roots of the resulting polynomial. - Use convexity and end behavior to verify the global minimum.

1. A general graph point is \((x, 4/x)\). Minimize the squared distance \(D(x)=(x+1)^2+\left(\frac{4}{x}+1\right)^2\). 2. Differentiate: \(D'(x)=2x+2-\frac{8}{x^2}-\frac{32}{x^3}\). Setting the derivative equal to zero and multiplying by \(x^3\) gives \(x^4+x^3-4x-16=0\). The positive solution \(x=2\) satisfies this equation. 3. The second derivative is \(D''(x)=2+\frac{16}{x^3}+\frac{96}{x^4}>0\) for \(x>0\). Thus \(D\) is strictly convex, so the critical point is unique. Since \(D(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), it is the absolute minimum. 4. The y-coordinate is \(f(2)=2\).

The closest point is \(Q(2, 2)\).

- Relate the triangle height, base, and inradius using a right-triangle cross section. - The height must be greater than twice the radius. - Minimize the square of the positive area to remove a square root. - Check the behavior at both ends of the domain.

1. Let \(b\) be the base length. The inradius geometry gives \(b^2=\frac{4r^2h}{h-2r}\), where \(h>2r\). With \(r=3\), \(b^2=\frac{36h}{h-6}\). 2. The triangle area is \(A=\frac{1}{2}bh\). Since \(A>0\), minimize its square: \(A^2=\frac{1}{4}b^2h^2=\frac{9h^3}{h-6}\). 3. Differentiate: \(\frac{d}{dh}(A^2)=\frac{18h^2(h-9)}{(h-6)^2}\). The only critical point in the domain is \(h=9\,\text{cm}\). 4. The derivative changes from negative to positive at \(h=9\). Also, \(A^2\to\infty\) as \(h\to6^+\) or \(h\to\infty\), so this is the absolute minimum.

The triangle area is minimized when \(h=9\,\text{cm}\).

- Use the positive x-intercept as the upper limit of integration. - Express the rectangle's dimensions in terms of \(u\) and \(a\). - Treat \(a\) as a constant when differentiating with respect to \(u\). - Set the maximum-area formula equal to the given value.

1. The positive x-intercept is \(x=\sqrt a\). Therefore, the area bounded by the graph and the axes is \(\int_0^{\sqrt a}(a-x^2)\, \mathrm{d}x\) \(=\left[ax-\frac13x^3\right]_0^{\sqrt a}\) \(=\frac23a\sqrt a\). 2. The rectangle has width \(u\) and height \(a-u^2\), so \(A(u)=u(a-u^2)=au-u^3\). Differentiate with respect to \(u\): \(A'(u)=a-3u^2\). The positive critical point is \(u=\sqrt{\frac a3}\). Since \(A''(u)=-6u<0\) and the area approaches \(0\) at both ends of the domain, this point gives the absolute maximum. 3. The maximum rectangle area is \(A_{\max}(a)=\sqrt{\frac a3}\left(a-\frac a3\right)\) \(=\frac{2a\sqrt a}{3\sqrt3}\). Set this equal to the given value: \(\frac{2a\sqrt a}{3\sqrt3}=\frac{16}{3\sqrt3}\). Thus, \(a^{3/2}=8\), so \(a=4\).

a) \(\frac23a\sqrt a\) b) \(A_{\max}(a)=\frac{2a\sqrt a}{3\sqrt3}\) c) \(a=4\)

- Rewrite the function to separate the parameter-dependent term. - Find x-values for which that parameter-dependent term is zero. - Use the x-intercepts as the limits of the area integral. - Differentiate the area function with respect to \(k\).

1. Rewrite the function as \(f_k(x)=x+\frac{4x-x^2}{k}\). A point is independent of \(k\) when \(4x-x^2=0\), so \(x=0\) or \(x=4\). The corresponding values are \(f_k(0)=0\) and \(f_k(4)=4\). Thus, all graphs pass through \((0, 0)\) and \((4, 4)\). 2. The x-intercepts of \(f_k\) are \(0\) and \(k+4\). The enclosed area is \(A(k)=\int_0^{k+4}\frac1k\big((k+4)x-x^2\big)\, \mathrm{d}x\) \(=\frac{(k+4)^3}{6k}\). 3. Differentiate: \(A'(k)=\frac{(k+4)^2(2k-4)}{6k^2}\). The only positive critical point is \(k=2\). The derivative changes from negative to positive there. Also, \(A(k)\to\infty\) as \(k\to0^+\) and as \(k\to\infty\), so this is the absolute minimum. 4. The minimum area is \(A(2)=\frac{6^3}{12}=18\) square units.

a) The common points are \((0, 0)\) and \((4, 4)\). b) The area is minimized when \(k=2\), and the minimum area is \(18\) square units.

- Evaluate both functions and their derivatives at \(x=0\). - Perpendicular tangent lines have slopes whose product is \(-1\). - Use improper integrals for the two unbounded parts of the region. - Use the orthogonality condition to eliminate one parameter before optimizing.

1. Since \(f_a(0)=g_b(0)=1\), the graphs intersect at \((0, 1)\). Their slopes there are \(f_a'(0)=a\) and \(g_b'(0)=b\). The graphs intersect orthogonally when the product of the slopes is \(-1\), so \(ab=-1\). 2. The enclosed area is the sum of two improper integrals: \(A(a, b)=\int_{-\infty}^0e^{ax}\, \mathrm{d}x+\int_0^\infty e^{bx}\, \mathrm{d}x\) \(=\frac1a-\frac1b\). 3. From \(ab=-1\), \(b=-\frac1a\). Substitute into the area formula: \(A(a)=\frac1a+a\). Then \(A'(a)=1-\frac1{a^2}\), so the positive critical point is \(a=1\). Since \(A''(a)=\frac2{a^3}>0\), this is the absolute minimum. Therefore, \(b=-1\), and the minimum area is \(A(1)=2\) square units.

1) The orthogonality condition is \(ab=-1\). 2) \(A(a, b)=\frac1a-\frac1b\) 3) \(a=1\), \(b=-1\), and the minimum area is \(2\) square units.

- Evaluate both functions and their derivatives at \(x=1\). - Integrate the upper function minus the lower function from \(1\) to \(e\). - Recall that an antiderivative of \(\ln x\) is \(x\ln x-x\). - Differentiate the area function with respect to \(k\).

1. At \(x=1\), \(f_k(1)=g_k(1)=0\). The slopes are \(f_k'(1)=k\) and \(g_k'(1)=-\frac1k\). Their product is \(-1\), so the graphs intersect orthogonally. 2. For \(1<x\le e\), \(f_k(x)>0\) and \(g_k(x)<0\). Thus, \(A(k)=\int_1^e\left(k\ln x+\frac{x-1}{k}\right)\, \mathrm{d}x\). Using \(\int\ln x\, \mathrm{d}x=x\ln x-x\), this becomes \(A(k)=k+\frac{(e-1)^2}{2k}\). 3. Differentiate: \(A'(k)=1-\frac{(e-1)^2}{2k^2}\). The positive critical point is \(k=\frac{e-1}{\sqrt2}\). Since \(A''(k)=\frac{(e-1)^2}{k^3}>0\), this critical point gives the absolute minimum.

1) The graphs meet at \((1, 0)\), and their tangent slopes have product \(-1\). 2) \(A(k)=k+\frac{(e-1)^2}{2k}\) 3) \(k=\frac{e-1}{\sqrt2}\)

- Find the zeros and determine where the graph lies below the x-axis. - Write the tangent line at \(x=a\), then set \(x=0\) to identify its y-intercept. - Differentiate the y-intercept function to find critical values. - Compare critical values and end behavior to identify the global maximum.

1. The zeros are \(x=-2\) and \(x=0\), and \(f(x)<0\) on \((-2, 0)\). Since an antiderivative is \(x^2e^x\), the area is \(A=-\int_{-2}^0(x^2+2x)e^x\,dx=-\left[x^2e^x\right]_{-2}^0=4e^{-2}\approx0.541\). 2. Differentiate: \(f'(x)=(x^2+4x+2)e^x\). The tangent line at \(x=a\) is \(y=(a^2+4a+2)e^a(x-a)+(a^2+2a)e^a\). 3. In slope-intercept form, \(t_a\) is \(y=(a^2+4a+2)e^a x+(-a^3-3a^2)e^a\). Thus the y-intercept is \(n(a)=(-a^3-3a^2)e^a\). 4. Differentiate: \(n'(a)=-a(a^2+6a+6)e^a\). The critical values are \(a=0\) and \(a=-3\pm\sqrt{3}\). 5. The value at \(a=-3-\sqrt{3}\) is positive, while \(n(0)=0\) and \(n(-3+\sqrt{3})<0\). Also, \(n(a)\to0\) as \(a\to-\infty\) and \(n(a)\to-\infty\) as \(a\to\infty\). Therefore, the global maximum occurs at \(a=-3-\sqrt{3}\).

a) \(A=4e^{-2}\approx0.541\) square units b) \(t_a: y=(a^2+4a+2)e^a x+(-a^3-3a^2)e^a\); the y-intercept is greatest at \(a=-3-\sqrt{3}\).

- Use an improper integral for the unbounded region. - Compare the growth of a polynomial with the decay of \(e^{-x}\) as \(x\to\infty\). - Set \(x=0\) in the tangent-line equation to obtain the y-intercept function. - Find and compare the critical values of that function.

1. The area is the improper integral \(A=\int_0^{\infty}x^2e^{-x}\,dx\). An antiderivative is \(-(x^2+2x+2)e^{-x}\). Since \((x^2+2x+2)e^{-x}\to0\) as \(x\to\infty\), \(A=0-(-2)=2\). 2. Differentiate: \(f'(x)=(2x-x^2)e^{-x}\). The tangent line at \(x=a\) is \(y=(2a-a^2)e^{-a}(x-a)+a^2e^{-a}\). 3. In slope-intercept form, the line is \(y=(2a-a^2)e^{-a}x+(a^3-a^2)e^{-a}\). Thus the y-intercept is \(n(a)=(a^3-a^2)e^{-a}\). 4. Differentiate: \(n'(a)=-a(a^2-4a+2)e^{-a}\). The critical values are \(a=0\) and \(a=2\pm\sqrt{2}\). 5. The value at \(a=2+\sqrt{2}\) is positive and greater than the values at the other critical points. Also, \(n(a)\to0\) as \(a\to\infty\) and \(n(a)\to-\infty\) as \(a\to-\infty\). Therefore, the global maximum occurs at \(a=2+\sqrt{2}\).

a) \(A=2\) square units b) \(t_a: y=(2a-a^2)e^{-a}x+(a^3-a^2)e^{-a}\); the y-intercept is greatest at \(a=2+\sqrt{2}\).

- Substitute the given time and amount into the model to solve for \(k\). - In part b), look for a factor that cancels from both sides. - Sketch points \(A\), \(B\), and \(C\); segment \(AB\) is horizontal. - Use the product rule to differentiate the area function. - Because \(a>5\), you may replace \(|a-5|\) with \(a-5\).

1. Use \(g_k(10)=500\): \(k(1-e^{-2})=500\). Thus \(k=\frac{500}{1-e^{-2}}\approx 578.26\). 2. To reach \(75\%\) of the saturation value, set \(g_k(t)=0.75k\): \(k(1-e^{-0.2t})=0.75k\). Since \(k>0\), divide by \(k\) to obtain \(e^{-0.2t}=0.25\). The equation no longer contains \(k\), so the time is independent of \(k\). 3. Solve for time: \(-0.2t=\ln(0.25)\), so \(t=5\ln(4)\approx 6.93\). 4. Since \(g_k'(0)=0.2k\), the tangent line at the origin is \(y=0.2kt\). It meets \(y=k\) when \(t=5\), so \(A=(5,k)\). Also, \(B=(a,k)\) and \(C=(a,k(1-e^{-0.2a}))\). 5. For \(a>5\), the base length is \(a-5\), and the vertical height is \(ke^{-0.2a}\). Therefore, \(F(a)=\frac{k}{2}(a-5)e^{-0.2a}\). 6. Differentiate: \(F'(a)=\frac{k}{2}e^{-0.2a}(2-0.2a)\). 7. Setting \(F'(a)=0\) gives \(a=10\). The derivative changes from positive to negative there, so the triangle has maximum area when \(a=10\).

a) \(k=\frac{500}{1-e^{-2}}\approx 578.26\) b) \(t=5\ln(4)\approx 6.93\) c) \(a=10\)

- Differentiate \(f\) twice before simplifying the difference. - Vertical distance is the absolute value of the difference of the function values. - Verify an antiderivative by differentiating it. - Determine the sign of \(f\) on the integration interval before computing geometric area.

1. Differentiate twice: \(f''(x)=(x^2+4x-1)e^x\). Therefore, \(d(x)=f''(x)-f(x)=(4x+2)e^x\). 2. Differentiate \(d\): \(d'(x)=(4x+6)e^x\). The only critical number is \(x=-\frac{3}{2}\). Since \(d''\left(-\frac{3}{2}\right)=4e^{-3/2}>0\), \(d\) has a local minimum there. Its value is \(d\left(-\frac{3}{2}\right)=-4e^{-3/2}\). 3. The actual vertical distance is \(|d(x)|\), which reaches its minimum value \(0\) where \(d(x)=0\). Solving \((4x+2)e^x=0\) gives \(x=-\frac{1}{2}\). 4. Differentiate \(F\): \(F'(x)=(2x-2)e^x+(x^2-2x-1)e^x=(x^2-3)e^x=f(x)\). Therefore, \(F\) is an antiderivative of \(f\). 5. On \([-1,1]\), \(x^2-3<0\) and \(e^x>0\), so \(f(x)<0\). Therefore, the area is \(-\int_{-1}^{1}f(x)\,dx=-[F(1)-F(-1)]=2e+\frac{2}{e}\approx6.172\) square units.

a) The signed difference \(d\) has a local minimum at \(\left(-\frac{3}{2},-4e^{-3/2}\right)\). The vertical distance is minimized at \(x=-\frac{1}{2}\), where it equals \(0\). b) \(F'(x)=f(x)\), so \(F\) is an antiderivative. c) \(2e+\frac{2}{e}\approx6.172\) square units.

- Use the distance formula for two points in the coordinate plane. - Minimizing a positive distance is equivalent to minimizing its square. - Differentiate the squared-distance function and find its critical points. - Use the derivative of \(f\) to find the tangent slope. - Two nonvertical lines are perpendicular when the product of their slopes is \(-1\).

1. The distance from \(Q(x, 0.5x^2)\) to \(P(6, 0)\) is \(d(x)=\sqrt{(x-6)^2+(0.5x^2)^2}\). Squaring and expanding gives \(d^2(x)=(x-6)^2+0.25x^4=0.25x^4+x^2-12x+36\). 2. Minimize \(g(x)=0.25x^4+x^2-12x+36\) on \(x\ge0\). Its derivative is \(g'(x)=x^3+2x-12=(x-2)(x^2+2x+6)\). The quadratic factor has no real zeros, so the only critical point is \(x=2\). Because \(g''(x)=3x^2+2>0\) for every \(x\), \(g\) is strictly convex and this critical point is the absolute minimum. Since \(f(2)=2\), the connection point is \(Q_E=(2, 2)\). 3. The slope of the tangent to the bypass at \(Q_E\) is \(f'(2)=2\). The slope of the access road is \(m=\frac{2-0}{2-6}=-\frac12\). Their product is \(2\cdot\left(-\frac12\right)=-1\), so the two lines are perpendicular.

a) \(d(x)=\sqrt{(x-6)^2+(0.5x^2)^2}\), and \(d^2(x)=0.25x^4+x^2-12x+36\). b) \(Q_E=(2, 2)\) c) The tangent slope is \(2\), and the access-road slope is \(-\frac12\). Their product is \(-1\), so the roads are perpendicular at \(Q_E\).

- Use the formula for the area of a right triangle. - Apply the product rule to the area function. - An exponential expression is never zero. - For the unbounded region, integrate to a variable upper bound and then take a limit. - Divide the maximum triangle area by the total area to find the percentage.

1. The triangle has base \(u\) and height \(f(u)=6e^{-0.4u}\). Therefore, \(A(u)=\frac12\cdot u\cdot6e^{-0.4u}=3ue^{-0.4u}\). 2. Differentiate using the product rule: \(A'(u)=(3-1.2u)e^{-0.4u}\). Since the exponential factor is positive, \(A'(u)=0\) when \(3-1.2u=0\), which gives \(u=2.5\). The derivative changes from positive to negative there, and the area approaches \(0\) at both ends of its domain, so this is the absolute maximum. The maximum area is \(A(2.5)=7.5e^{-1}=\frac{7.5}{e}\approx2.76\) square units. 3. The total area is \(A_{\text{total}}=\int_0^\infty6e^{-0.4x}\, \mathrm{d}x\) \(=\lim_{b\to\infty}\left[-15e^{-0.4x}\right]_0^b\) \(=\lim_{b\to\infty}\left(15-15e^{-0.4b}\right)=15\). 4. The requested percentage is \(\frac{7.5e^{-1}}{15}\cdot100\%=\frac{50}{e}\%\approx18.4\%\).

a) \(A(u)=3ue^{-0.4u}\) b) \(u=2.5\), and the maximum area is \(\frac{7.5}{e}\approx2.76\) square units. c) \(A_{\text{total}}=15\) square units. d) The maximum triangle area is approximately \(18.4\%\) of the total area.

- Use the distance formula between \(P\) and a general point \(Q\) on the graph. - To minimize a positive square root, minimize its radicand. - Sketch the enclosed region and separate it into an area under the curve and a triangle. - Use the power rule for integration.

1. By the distance formula, \(d(x)=\sqrt{(x-3)^2+(2\sqrt{x})^2}=\sqrt{x^2-2x+9}\). 2. Since the square-root function is increasing, minimizing \(d(x)\) is equivalent to minimizing \(h(x)=x^2-2x+9\) for \(x\ge0\). 3. Because \(h^{\prime}(x)=2x-2\), the only critical point is \(x=1\). Also, \(h^{\prime\prime}(x)=2>0\), so this point gives the minimum. Then \(f(1)=2\), and \(Q_E=(1, 2)\). 4. From \(x=0\) to \(x=1\), the enclosed area is \(A_1=\int_0^1 2\sqrt{x}\,dx=\left[\frac{4}{3}x^{3/2}\right]_0^1=\frac{4}{3}\). 5. From \(x=1\) to \(x=3\), the area below \(\overline{PQ_E}\) is a right triangle with base \(2\) and height \(2\), so \(A_2=\frac{1}{2}\cdot2\cdot2=2\). 6. The total area is \(A_1+A_2=\frac{4}{3}+2=\frac{10}{3}\) square units.

a) \(d(x)=\sqrt{x^2-2x+9}\) b) \(Q_E=(1, 2)\) c) \(\frac{10}{3}\) square units

- Write the area formula for each trapezoid. - Evaluate \(g\) at the endpoints of the interval. - Add and simplify the two trapezoid areas before differentiating. - Compare the maximum approximation with the exact definite integral.

1. The two trapezoids have areas \(A_1=\frac{g(0)+g(u)}{2}u\) and \(A_2=\frac{g(u)+g(4)}{2}(4-u)\). Since \(g(0)=2\) and \(g(4)=6\), their total area simplifies to \(A(u)=2g(u)-2u+12\) \(=4\sqrt{u}-2u+16\). 2. Differentiate: \(A'(u)=\frac{2}{\sqrt{u}}-2\). Setting the derivative equal to zero gives \(\sqrt{u}=1\), so \(u=1\). Because \(A''(u)=-u^{-3/2}<0\) on \((0, 4)\), this critical point gives the absolute maximum. 3. Since \(g(1)=4\), the point is \(R=(1, 4)\), and the maximum pentagon area is \(A(1)=18\). 4. The exact area is \(\int_0^4(2\sqrt{x}+2)\, \mathrm{d}x\) \(=\left[\frac43x^{3/2}+2x\right]_0^4=\frac{56}{3}\). Therefore, the percent error is \(\frac{\frac{56}{3}-18}{\frac{56}{3}}\cdot100\%=\frac{25}{7}\%\approx3.57\%\).

a) The pentagon's area is greatest when \(u=1\), and \(R=(1, 4)\). The maximum approximation is \(18\) square units. b) The percent error is \(\frac{25}{7}\%\approx3.57\%\).

- Imagine translating line \(g\) parallel to itself toward the parabola. - Match the slope of the parabola's tangent to the slope of \(g\). - Use the derivative of \(f\) to find the point of tangency. - Use the point-to-line distance formula for the actual distance.

1. At the minimum distance, the shortest segment from the line to the parabola is perpendicular to the line. It is also normal to the parabola at the closest point, so it is perpendicular to the tangent there. Therefore, the tangent and line \(g\) are parallel. 2. The derivative of the parabola is \(f'(x)=0.5x\). Line \(g\) has slope \(0.5\), so parallelism requires \(0.5x=0.5\). Thus, \(x=1\), and \(f(1)=2.25\). Therefore, \(P=(1, 2.25)\). 3. Write the line as \(0.5x-y-1=0\). The point-to-line distance is \(d=\frac{|0.5(1)-2.25-1|}{\sqrt{0.5^2+(-1)^2}}\) \(=\frac{2.75}{\sqrt{1.25}}=\frac{11}{2\sqrt5}\approx2.46\) units.

a) The shortest connecting segment is perpendicular to both line \(g\) and the tangent at the closest point, so the tangent and \(g\) are parallel. b) \(P=(1, 2.25)\), and the minimum distance is \(\frac{11}{2\sqrt5}\approx2.46\) units.