An energy company models a family of annual plans by \(K_p(x)=px+(300-2000p)\), where \(x\) is annual usage in kWh, \(p\) is the price per kWh in dollars, and \(K_p(x)\) is annual cost in dollars.
1. For a plan with \(p=0.12\), identify the slope and y-intercept and interpret both in context.
2. Show that all graphs in the family pass through one common point. Find the point and interpret it in context.
3. A customer uses \(3500\,\text{kWh}\) per year. Determine whether a higher or lower value of \(p\) gives this customer a lower annual cost.
Hints
- Substitute the specified value of \(p\) before interpreting the slope and intercept.
- Rewrite the family so the coefficient of \(p\) is easy to see.
- Find the usage value that makes the \(p\)-dependent part equal to \(0\).
- For the final comparison, substitute \(3500\) and see how the cost changes as \(p\) changes.
Solution
1. With \(p=0.12\), \(K_{0.12}(x)=0.12x+60\). The slope \(0.12\) means each additional kWh adds \(\$0.12\) to the modeled annual cost. The y-intercept \(60\) means the model gives an annual cost of \(\$60\) at zero usage.
2. Rewrite the family as \(K_p(x)=p(x-2000)+300\). At \(x=2000\), the term containing \(p\) is \(0\), so every plan gives \(K_p(2000)=300\). The common point is \((2000,300)\): every plan costs \(\$300\) at \(2000\,\text{kWh}\).
3. At \(x=3500\), \(K_p(3500)=300+1500p\). This increases as \(p\) increases, so a lower value of \(p\) gives the lower annual cost.
Answer
1. For \(p=0.12\), slope \(=0.12\) dollars per kWh and y-intercept \(=60\) dollars. The slope is the added cost per additional kWh; the intercept is the modeled cost at zero usage.
2. Common point: \((2000,300)\). At \(2000\,\text{kWh}\), every plan costs \(\$300\).
3. A lower value of \(p\) gives the lower annual cost at \(3500\,\text{kWh}\).