Chapter 8, Problem 55RE

(a)

To determine

To find number of people who entered park by 5:00pm $\left(t=17\right)$

Answer to Problem 55RE

Number of people who entered park by 5:00pm $\left(t=17\right)$ $\approx 6004$

Explanation of Solution

Given:

The rate at which people enter an amusement park on a given day is modelled by the function E is defined by

  $E\left(t\right)=\frac{15600}{\left(t^2-24t+160\right)}$

  $E\left(t\right)$ is measured in people per hour and time t is measured in hours after midnight and function is valid for $9\le t\le 23$ ,the hours during which park is open.

At $t=9$ , there are no people in park.

Concept Used:

Application of integrals

Calculation:

Number of people who entered park by 5:00pm $\left(t=17\right)$ = ${\displaystyle \underset{9}{\overset{17}{\int }}E\left(t\right)dt}$

  $\begin{array}{l}{\displaystyle \underset{9}{\overset{17}{\int }}E\left(t\right)dt}\\ ={\displaystyle \underset{9}{\overset{17}{\int }}\left(\frac{15600}{\left(t^2-24t+160\right)}\right)}dt\\ =15600{\displaystyle \underset{9}{\overset{17}{\int }}\left(\frac{1}{\left(t^2-24t+160\right)}\right)}dt\\ =15600{\displaystyle \underset{9}{\overset{17}{\int }}\left(\frac{1}{\left({\left(t-12\right)}^2+16\right)}\right)}dt\left[complete\text{ the square}\right]\\ =15600{\displaystyle \underset{9}{\overset{17}{\int }}\left(\frac{1}{16\left({\left(\frac{t-12}{4}\right)}^2+1\right)}\right)}dt\\ \text{=}\frac{15600}{16}{\displaystyle \underset{9}{\overset{17}{\int }}\left(\frac{1}{\left({\left(\frac{t-12}{4}\right)}^2+1\right)}\right)}dt\\ =975\times 4\left[\left({\tan }^{-1}\left(\frac{t-12}{4}\right)\right){|}_9^{17}\right]\\ =3900\left[\left({\tan }^{-1}\left(\frac{17-12}{4}\right)-{\tan }^{-1}\left(\frac{9-12}{4}\right)\right)\right]\\ =3900\left({\tan }^{-1}\left(\frac{5}{4}\right)-{\tan }^{-1}\left(\frac{-3}{4}\right)\right)\\ =3900\left({\tan }^{-1}\left(\frac{5}{4}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)\right)\\ \approx 6004.270\approx 6004\end{array}$

Conclusion:

Number of people who entered park by 5:00pm $\left(t=17\right)$ $\approx 6004$

(b)

To determine

To find number of dollars collected from admissions to the park on the given day

Answer to Problem 55RE

Number of dollars collected from admissions to the park on the given day=$104,041

Explanation of Solution

Given:

Price of admission to the park until 5:00pm $\left(P_1\right)$ =$15

Price of admission to the park after 5:00pm $\left(P_2\right)$ =$11

Concept Used:

Number of dollars collected from admissions to the park on the given day = number of people (who enter park by 5:00 pm) $\times$

  $\left(P_1\right)$ + number of people (who enter park between 5:00 pm and 11:00 pm) $\times$

  $\left(P_2\right)$

Calculation:

Number of people who entered park by 5:00pm $\left(t=17\right)$

  $\approx 6004$ [from part (a)]

Number of people who entered park between 5:00 pm and 11:00 pm= ${\displaystyle \underset{17}{\overset{23}{\int }}E\left(t\right)dt}$

  $\begin{array}{l}{\displaystyle \underset{17}{\overset{23}{\int }}E\left(t\right)dt}\\ ={\displaystyle \underset{17}{\overset{23}{\int }}\left(\frac{15600}{\left(t^2-24t+160\right)}\right)}dt\\ =15600{\displaystyle \underset{17}{\overset{23}{\int }}\left(\frac{1}{\left(t^2-24t+160\right)}\right)}dt\\ =15600{\displaystyle \underset{17}{\overset{23}{\int }}\left(\frac{1}{\left({\left(t-12\right)}^2+16\right)}\right)}dt\left[complete\text{ the square}\right]\\ =15600{\displaystyle \underset{17}{\overset{23}{\int }}\left(\frac{1}{16\left({\left(\frac{t-12}{4}\right)}^2+1\right)}\right)}dt\\ \text{=}\frac{15600}{16}{\displaystyle \underset{17}{\overset{23}{\int }}\left(\frac{1}{\left({\left(\frac{t-12}{4}\right)}^2+1\right)}\right)}dt\\ =975\times 4\left[\left({\tan }^{-1}\left(\frac{t-12}{4}\right)\right){|}_{17}^{23}\right]\\ =3900\left[\left({\tan }^{-1}\left(\frac{23-12}{4}\right)-{\tan }^{-1}\left(\frac{17-12}{4}\right)\right)\right]\\ =3900\left({\tan }^{-1}\left(\frac{11}{4}\right)-{\tan }^{-1}\left(\frac{5}{4}\right)\right)\\ =3900\left({\tan }^{-1}\left(\frac{5}{4}\right)-{\tan }^{-1}\left(\frac{3}{4}\right)\right)\\ =1271.283\approx 1271\end{array}$

Number of dollars collected from admissions to the park on the given day

  $\begin{array}{l}=\$15\times 6004+\$11\times 1271\\ =\$90,060+\$13,981\\ =\$104,041\end{array}$

Conclusion:

Number of dollars collected from admissions to the park on the given day=$104,041

(c)

To determine

To find the value of $H^{\prime }\left(17\right)$ and explain the meaning of $H\left(17\right)$ and $H^{\prime }\left(17\right)$ in context to the park

Answer to Problem 55RE

Value of $H^{\prime }\left(17\right)$ = $-380.281$

Explanation of Solution

Given:

  $H\left(t\right)={\displaystyle \underset{9}{\overset{t}{\int }}\left(E\left(x\right)-L\left(x\right)\right)dx}\text{ for 9}\le \text{t}\le \text{23}$

The value of $H\left(17\right)$ to the nearest whole number is 3725

Concept Used:

  $H\left(17\right)$ Means that there were 3725 people in the park at $\left(t=17\right)$

As,

  $H\left(t\right)={\displaystyle \underset{9}{\overset{t}{\int }}\left(E\left(x\right)-L\left(x\right)\right)dx}\text{ for 9}\le \text{t}\le \text{23}$

  $H^{\prime }\left(t\right)=E\left(t\right)-L\left(t\right)$

Calculation:

For t=7

  $H^{\prime }\left(17\right)=E\left(17\right)-L\left(17\right)$

  $\begin{array}{l}E\left(17\right)-L\left(17\right)\\ =\frac{15600}{\left({\left(17\right)}^2-24\left(17\right)+160\right)}-\frac{9890}{\left({\left(17\right)}^2-38\left(17\right)+370\right)}\\ =\frac{15600}{41}-\frac{9890}{13}\end{array}$

  $\begin{array}{l}=380.488-760.769\\ =-380.281\end{array}$

Conclusion:

  $H\left(17\right)$ Means that there were 3725 people in the park at $\left(t=17\right)$

As $H^{\prime }\left(17\right)$ = $-380.281$

  $H^{\prime }\left(17\right)$ Means that the number of people in the park was decreasing roughly by 380 people per hour at time $\left(t=17\right)$

(c)

To determine

To find time at which number of people in the park is maximum

Answer to Problem 55RE

At t=15.795 number of people in the park is maximum

Explanation of Solution

Given:

Time t, for $9\le t\le 23$

Concept Used:

To find the maximum people in the park using $H^{\prime }\left(t\right)=0$

Calculation:

  $\begin{array}{l}{H}^{\prime }\left(t\right)=0\\ \Rightarrow E\left(t\right)-L\left(t\right)=0\\ \Rightarrow E\left(t\right)=L\left(t\right)\\ \Rightarrow \frac{15600}{\left(t^2-24t+160\right)}=\frac{9890}{\left(t^2-38t+370\right)}\end{array}$

Solving the equation by using graphing utility to get value of t

  

Conclusion:

From graph,

At t=15.795 number of people in the park is maximum