I am easy to waste and unstoppable What am I ??

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

Arlecino [84]

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 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:

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

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.

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

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