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Daily website visits during a campaign are modeled by \(S_k(t)=k(20-t)e^{-0.1t}+10k\), where \(k>0\) and \(t\) is measured in days.
a) Find the long-term number of visits in terms of \(k\).
b) Campaign 1 begins at \(t=0\) with \(k=100\). Campaign 2 begins at \(t=50\) with \(k=150\). Write a model for the combined daily visits \(G(t)\) for \(t\geq50\).
c) Find the combined daily visits at \(t=60\).
d) Find the long-term combined daily visits from both campaigns.
Hints
- Find the limit of each time-dependent term.
- Shift the second campaign's time input by its start time.
- Add the two campaign models.
- Add the long-term values of the individual campaigns.
Solution
1. As \(t\to\infty\), \((20-t)e^{-0.1t}\to0\), so \(S_k(t)\to10k\).
2. For \(t\geq50\), \(G(t)=S_{100}(t)+S_{150}(t-50)\). Therefore, \(G(t)=100(20-t)e^{-0.1t}+150(70-t)e^{-0.1(t-50)}+2500\).
3. \(G(60)=-4000e^{-6}+1500e^{-1}+2500\approx3041.90\), so the model predicts about \(3042\) visits.
4. The long-term total is \(10(100)+10(150)=2500\) visits.
Answer
a) \(10k\)
b) \(G(t)=100(20-t)e^{-0.1t}+150(70-t)e^{-0.1(t-50)}+2500\), for \(t\geq50\)
c) About \(3042\) visits
d) \(2500\) visits
