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Veronika [31]
2 years ago
14

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)​

Advanced Placement (AP)
1 answer:
Arlecino [84]2 years ago
8 0

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(\frac{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(\frac{t}{5})} \, dt - \int\limits^{20}_{10} {12e^{\frac{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:

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

Evaluate at t = 20:


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

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

This means that at t = 20, there is a <u>NET DECREASE</u> 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(\frac{t}{5}) - 12e^{\frac{t}{10}} - 20 = 0\\\\60 - 12(cos(\frac{t}{5}) + e^{\frac{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.

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

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