the number of visitors p to a website in a given week over a 1 - year period is given by ( p(t)=121+(t…

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 ).\na) over what interval of time during the 1 - year period is the number of visitors decreasing?\nb) over what interval of time during the 1 - year period is the number of visitors increasing?\nc) find the critical point, and interpret its meaning.\na) the number of visitors is decreasing over the interval \n(simplify your answer. type integers or decimals rounded to three decimal places as needed. type your answer in interval notation.)

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 ).\na) over what interval of time during the 1 - year period is the number of visitors decreasing?\nb) over what interval of time during the 1 - year period is the number of visitors increasing?\nc) find the critical point, and interpret its meaning.\na) the number of visitors is decreasing over the interval \n(simplify your answer. type integers or decimals rounded to three decimal places as needed. type your answer in interval notation.)

Answer

Explanation:

Step1: Differentiate the function

Differentiate (P(t)=121+(t - 88)e^{0.02t}) using the product rule ((uv)^\prime=u^\prime v+uv^\prime), where (u=t - 88), (u^\prime=1), (v = e^{0.02t}), (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))). Simplify (1+0.02(t - 88)=1+0.02t-1.76 = 0.02t - 0.76). So (P^\prime(t)=e^{0.02t}(0.02t - 0.76)). Since (e^{0.02t}>0) for all (t), the sign of (P^\prime(t)) is determined by (0.02t - 0.76).

Step2: Find when (P^\prime(t)<0) (for decreasing)

Set (0.02t - 0.76<0). Solve for (t): (0.02t<0.76), (t < \frac{0.76}{0.02}=38). Since (1\leq t\leq52), the function is decreasing when (1\leq t<38).

Step3: Find when (P^\prime(t)>0) (for increasing)

Set (0.02t - 0.76>0). Solve for (t): (0.02t>0.76), (t>\frac{0.76}{0.02} = 38). Since (1\leq t\leq52), the function is increasing when (38<t\leq52).

Step4: Find the critical point

Set (P^\prime(t) = 0). Since (e^{0.02t}\neq0), set (0.02t - 0.76 = 0), (t = 38). When (t = 38), (P(38)=121+(38 - 88)e^{0.02\times38}=121-50e^{0.76}\approx121 - 50\times2.138=121-106.9 = 14.1). The critical point ((38,14.1)) is a local minimum. It means that in the 38 - th week, the number of website visitors reaches the lowest value within the 1 - year period.

Answer:

a) The number of visitors is decreasing over the interval ([1,38)) b) The number of visitors is increasing over the interval ((38,52]) c) The critical point is (t = 38). It is a local minimum, representing the week ((t = 38)) when the number of website visitors is the lowest within the 1 - year period.