Question

the number of visitors ( p ) to a website in a given week over a 1 - year period is given by ( p(t)=121+(t - 88)e^{0.02t} ), where ( t ) is the week and ( 1leq tleq52 ).
a) over what interval of time during the 1 - year period is the number of visitors decreasing?
b) over what interval of time during the 1 - year period is the number of visitors increasing?
c) find the critical point, and interpret its meaning.
a) the number of visitors is decreasing over the interval ( (1,38) ).
(simplify your answer. type integers or decimals rounded to three decimal places as needed. type your answer in interval notation.)
b) the number of visitors is increasing over the interval ( (38,52) ).
(simplify your answer. type integers or decimals rounded to three decimal places as needed. type your answer in interval notation.)
c) the critical point is ( (38,14.086) ).
(type an ordered pair. type integers or decimals rounded to three decimal places as needed.)
interpret what the critical point means.
the critical point means that the number of visitors was ( ) in week ( ) at approximately ( ) visitors.
(round to the nearest integer as needed.)

Explanation:

Step1: Analyze the derivative of \(P(t)\)

First, find the derivative of \(P(t)=121+(t - 88)e^{0.02t}\) using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u=t - 88\) and \(v = e^{0.02t}\).
\(u^\prime=1\), \(v^\prime=0.02e^{0.02t}\)
\(P^\prime(t)=e^{0.02t}+0.02(t - 88)e^{0.02t}=e^{0.02t}(1 + 0.02(t - 88))=e^{0.02t}(0.02t-1.76 + 1)=e^{0.02t}(0.02t - 0.76)\)

Step2: Find the critical point

Set \(P^\prime(t)=0\). Since \(e^{0.02t}>0\) for all \(t\), we solve \(0.02t-0.76 = 0\)
\(0.02t=0.76\), so \(t=\frac{0.76}{0.02}=38\)

Step3: Determine the intervals of increase and decrease

• For \(t\in(1,38)\), let's take a test - value, say \(t = 20\). Then \(P^\prime(20)=e^{0.02\times20}(0.02\times20 - 0.76)=e^{0.4}(0.4 - 0.76)<0\). So \(P(t)\) is decreasing on \((1,38)\)
• For \(t\in(38,52)\), let's take a test - value, say \(t = 40\). Then \(P^\prime(40)=e^{0.02\times40}(0.02\times40 - 0.76)=e^{0.8}(0.8 - 0.76)>0\). So \(P(t)\) is increasing on \((38,52)\)

Step4: Interpret the critical point

When \(t = 38\), \(P(38)=121+(38 - 88)e^{0.02\times38}=121-50e^{0.76}\)
\(e^{0.76}\approx2.138\), so \(P(38)=121-50\times2.138=121 - 106.9\approx14\)

Answer:

The critical point means that the number of visitors was \(\text{minimum}\) in week \(38\) at approximately \(14\) visitors.