Question

1. The rate at which people enter a movie theater on a given day is modeled by the function S defined by S(t) = 80 -12 cos 6 The rate at which people leave the same movie theater is modeled by the function R defined by r(t) = 12eKo + 20. Both S (t) and r(t) are measured in people per hour and these functions are t valid for 10 < t < 22. At time t = 10, there are no people in the movie theater.(Problems are in the Picture and I need help)​

Answer 1

Hi there!

a.

To find the total amount of people that have ENTERED by t = 20, we must take the integral of the appropriate function.

`\text{Amount that entered} = \int\limits^(20)_(10) {S(t)} \, dt \\\\ = \int\limits^(20)_(10) {80 - 12cos((t)/(5))} \, dt`

Evaluate using a calculator:

`= 899.97 \approx \boxed{900\text{ people}}`

b.

To solve, we can find the total amount of people that have entered of the interval and subtract the total amount of people that have left from this value.

In other terms:

`\text{Amount of people} = \int\limits^(20)_(10) {S(t)} \, dt - \int\limits^(20)_(10) {R(t)} \, dt`

We can evaluate using a calculator (math-9 on T1-84):

`\text{\# of people} = \int\limits^(20)_(10) {80-12cos((t)/(5))} \, dt - \int\limits^(20)_(10) {12e^{(t)/(10)}+20} \, dt`

`= 899.97 - 760.49 = 139.47 \approx \boxed{139 \text{ people}}`

c.

If:

`P(t) = \int\limits^t_(10) {S(t) - R(t)} \, dt`

Then:

`(dP)/(dt) = P'(t)= (d)/(dt)\int\limits^t_(10) {S(t) - R(t)} \, dt = S(t) - R(t)`

Evaluate at t = 20:

`S(20) = 80 - 12cos((20)/(5)) = 87.844\\\\R(20) = 12e^{(20)/(10)} + 20 = 108.669`

`S(20) - R(20) = 87.844 - 108.669 = -20.823`

This means that at t = 20, there is a NET DECREASE of people at the movie theater of around 20.823 (21) people per hour.

d.

To find the maximum, we must use the first-derivative test.

Set S(t) - R(t) equal to 0:

`80 - 12cos((t)/(5)) - 12e^{(t)/(10)} - 20 = 0\\\\60 - 12(cos((t)/(5)) + e^{(t)/(10)})= 0`

Graph the function with a graphing calculator and set the function equal to y = 0:

According to the graph, the graph of the first derivative changes from POSITIVE to NEGATIVE at t ≈ 17.78 hours, so there is a MAXIMUM at this value.

Thus, at t = 17.78 hours, the amount of people at the movie theater is a MAXIMUM.