Chapter 7.1, Problem 44E

a.

To determine

To find: The piece-wise function that describes the wait time of a person who arrives at time $T$ .

Answer to Problem 44E

The wait time of a person in minutes who arrives at time $T$ is $W\left(T\right)=\left\{\begin{array}{ll}20+20T-5T^2,\hfill & 0\le T<2\hfill \\ 80-40T+10T^2,\hfill & 2\le T<3\hfill \\ -190+140T-20T^2,\hfill & 3\le T<4\hfill \\ 290-100T+10T^2\hfill & 4\le T\le 5\hfill \end{array}\right.$ .

Explanation of Solution

Given information:

The piece-wise function for the number of people waiting in the line from noon until 5 pm is,

  $L\left(T\right)=\left\{\begin{array}{ll}500+500T-125T^2,\hfill & 0\le T<2\hfill \\ 2000-1000T+250T^2,\hfill & 2\le T<3\hfill \\ -4750+3500T-500T^2,\hfill & 3\le T<4\hfill \\ 7250-2500T+250T^2\hfill & 4\le T\le 5\hfill \end{array}\right.$ .

Calculation:

Since the ride can 1500 per hour, the number of people taking the ride per minute is $\frac{1500}{60}=25$ .

If a person arrives at time $T$ , then the person will be behind $L\left(T\right)$ number of people in the line. The time it takes for $L\left(T\right)$ people to take the ride is $\frac{L\left(T\right)}{25}$ minutes. Therefore, the piece wise function that describes the wait time in minutes for a function arrives at time $T$ can be given as

  $W\left(T\right)=\left\{\begin{array}{ll}\frac{1}{25}\left(500+500T-125T^2\right),\hfill & 0\le T<2\hfill \\ \frac{1}{25}\left(2000-1000T+250T^2\right),\hfill & 2\le T<3\hfill \\ \frac{1}{25}\left(-4750+3500T-500T^2\right),\hfill & 3\le T<4\hfill \\ \frac{1}{25}\left(7250-2500T+250T^2\right),\hfill & 4\le T\le 5\hfill \end{array}\right.$ or $W\left(T\right)=\left\{\begin{array}{ll}20+20T-5T^2,\hfill & 0\le T<2\hfill \\ 80-40T+10T^2,\hfill & 2\le T<3\hfill \\ -190+140T-20T^2,\hfill & 3\le T<4\hfill \\ 290-100T+10T^2\hfill & 4\le T\le 5\hfill \end{array}\right.$ .

Conclusion:

The wait time of a person in minutes who arrives at time $T$ is $W\left(T\right)=\left\{\begin{array}{ll}20+20T-5T^2,\hfill & 0\le T<2\hfill \\ 80-40T+10T^2,\hfill & 2\le T<3\hfill \\ -190+140T-20T^2,\hfill & 3\le T<4\hfill \\ 290-100T+10T^2\hfill & 4\le T\le 5\hfill \end{array}\right.$ .

b.

To determine

To find: The average wait time between noon and 5 pm.

Answer to Problem 44E

The wait time of a person in minutes is $\frac{124}{3}$ minutes.

Explanation of Solution

Given information:

The piece-wise function for the number of people waiting in the line from noon until 5 pm is,

  $L\left(T\right)=\left\{\begin{array}{ll}500+500T-125T^2,\hfill & 0\le T<2\hfill \\ 2000-1000T+250T^2,\hfill & 2\le T<3\hfill \\ -4750+3500T-500T^2,\hfill & 3\le T<4\hfill \\ 7250-2500T+250T^2\hfill & 4\le T\le 5\hfill \end{array}\right.$ .

Calculation:

From part (a), the wait time of a person in minutes who arrives at time $T$ is $W\left(T\right)=\left\{\begin{array}{ll}20+20T-5T^2,\hfill & 0\le T<2\hfill \\ 80-40T+10T^2,\hfill & 2\le T<3\hfill \\ -190+140T-20T^2,\hfill & 3\le T<4\hfill \\ 290-100T+10T^2\hfill & 4\le T\le 5\hfill \end{array}\right.$ .

So, the average wait time between noon and 5 pm is given by $W_{\text{Average}}=\frac{1}{5-0}{\displaystyle \underset{0}{\overset{5}{\int }}W\left(T\right)dT}$ .

Now, evaluate the integral ${\displaystyle \underset{0}{\overset{5}{\int }}W\left(T\right)dT}$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{5}{\int }}W\left(T\right)dT}={\displaystyle \underset{0}{\overset{2}{\int }}W\left(T\right)dT}+{\displaystyle \underset{2}{\overset{3}{\int }}W\left(T\right)dT}+{\displaystyle \underset{3}{\overset{4}{\int }}W\left(T\right)dT}+{\displaystyle \underset{4}{\overset{5}{\int }}W\left(T\right)dT}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}\left(20+20T-5T^2\right)dT}+{\displaystyle \underset{2}{\overset{3}{\int }}\left(80-40T+10T^2\right)dT}+{\displaystyle \underset{3}{\overset{4}{\int }}\left(-190+140T-20T^2\right)dT}\\ +{\displaystyle \underset{4}{\overset{5}{\int }}\left(290-100T+10T^2\right)dT}\\ =\frac{200}{3}+\frac{130}{3}+\frac{160}{3}+\frac{130}{3}\\ =\frac{620}{3}\end{array}$

Therefore, the average time in minutes is $W_{\text{Average}}=\frac{1}{5}\left(\frac{620}{3}\right)=\frac{124}{3}$ .

Conclusion:

The wait time of a person in minutes is $\frac{124}{3}$ minutes.