c. During the hours that the park is open, 9≤t≤23, what is the maximum number of people in the park at any given moment? Justify your answer. The rate at which people enter an amusement park on a given day is modeled by the function E defined by15600E(t):?-24t +160The rate at which people leave the same amusement part on the same day is modeled by the function Ldefined by9890L(t) =1? - 38t + 370Both E(t) and L(t) are measured in people per hour and time t is measured in number of hours aftermidnight. These functions are valid for 9 b. Let H(t) = (E(1) – L(t)dt for 9

Consider the given: a) The price of admission to the park is $15 until 5:00 PM.At 5 p.m.,…

Answered: The rate at which people enter an…

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter3: Functions

Section3.2: Domain And Range

Problem 61SE: The cost in dollars of making x items is given by the function Cx)=10x+500. a. The fixed cost is...

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c. During the hours that the park is open, 9≤t≤23, what is the maximum number of people in the park at any given moment? Justify your answer.

The rate at which people enter an amusement park on a given day is modeled by the function E defined by
15600
E(t):
?-24t +160
The rate at which people leave the same amusement part on the same day is modeled by the function L
defined by
9890
L(t) =
1? - 38t + 370
Both E(t) and L(t) are measured in people per hour and time t is measured in number of hours after
midnight. These functions are valid for 9 <I< 23, the hours during which the park is open. At t = 9,
there are no people in the park.
a. The price of admission to the park is $15 until 5:00-p.m. After 5:00 p.m., the price of admission to
the park is S11. How many dollars are collected from admissions to the park on the given day? Round
your answer to the nearest whole number.

b. Let H(t) = (E(1) – L(t)dt for 9 <t5 23. Find the values of H(17) and H'(17) and explain, using
correct units, the meaning of both values in the context of the park.

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Consider the given:

$\begin{array}{l} The price of admission to the park is $15 until 5:00 PM. \\ At 5 p.m., t = 17 \\ S o, \\ P = 15 \int_{9}^{17} \frac{15600}{t^{2} - 24 t + 160} d t + 11 \int_{17}^{23} \frac{15600}{t^{2} - 24 t + 160} d t \\ Let u = t - 12 \\ P = 15 \int_{- 3}^{5} \frac{15600}{(u^{2} + 16)} d u + 11 \int_{5}^{11} \frac{15600}{(u^{2} + 16)} d u \\ P = 15 {[15600 \tan^{- 1} (\frac{u}{4})]}_{- 3}^{5} + 11 {[15600 \tan^{- 1} (\frac{u}{4})]}_{5}^{11} \\ P = 15600 \tan^{- 1} (\frac{5}{4}) + 58500 \tan^{- 1} (\frac{3}{4}) + 42900 \tan^{- 1} (\frac{11}{4}) \\ P = 104048.165 \\ P = $ 104048 \end{array}$

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