Chapter 7.1, Problem 43E

To determine

To find: The piece-wise function $L\left(T\right)$ that describes the total number of people waiting in the line at any time $T$ .

Answer to Problem 43E

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

Explanation of Solution

Given information:

The rate at which people joining the line from noon until 5 pm is given by the piece-wise function,

  $R(t)=\left\{\begin{array}{ll}2000-250t,\hfill & 0\le t<2\hfill \\ 500+500t,\hfill & 2\le t<3\hfill \\ 5000-1000t,\hfill & 3\le t<4\hfill \\ -1000+500t,\hfill & 4\le t\le 5\hfill \end{array}\right.$ .

Calculation:

The rate at which people are waiting in the line at any time can be given by the function, $W(t)=\left\{\begin{array}{ll}2000-250t-1500,\hfill & 0\le t<2\hfill \\ 500+500t-1500,\hfill & 2\le t<3\hfill \\ 5000-1000t-1500,\hfill & 3\le t<4\hfill \\ -1000+500t-1500,\hfill & 4\le t\le 5\hfill \end{array}\right.$ .

Now, the total number of people waiting at any time $T$ from noon till 2 pm is,

  $\begin{array}{c}L\left(T\right)=500+{\displaystyle \underset{0}{\overset{T}{\int }}\left(2000-250t-1500\right)dt}\\ =500+500T-125T^2\end{array}$

At $T=2$ , $L\left(2\right)=500+500\times 2-125{\left(2\right)}^2=1000$ . That means, 1000 people are waiting in the line at 2 pm.

The function giving the total number of people waiting in the line from 2 pm to 3 pm is,

  $\begin{array}{c}L\left(T\right)=1000+{\displaystyle \underset{2}{\overset{T}{\int }}\left(500+500t-1500\right)dt}\\ L\left(T\right)=2000-1000T+250T^2\end{array}$

At $T=3$ , $L\left(3\right)=2000-1000\left(3\right)+250{\left(3\right)}^2=1250$ . That means, 1250 people are waiting in the line at 3 pm.

The function giving the total number of people waiting in the line from 3 pm to 4 pm is,

  $\begin{array}{c}L\left(T\right)=1250+{\displaystyle \underset{3}{\overset{T}{\int }}\left(5000-1000t-1500\right)dt}\\ L\left(T\right)=-4750+3500T-500T^2\end{array}$

At $T=4$ , $L\left(4\right)=-4750+3500\left(4\right)-500{\left(4\right)}^2=1250$ . That means, 1250 people are waiting in the line at 4 pm.

The function giving the total number of people waiting in the line from 3 pm to 4 pm is,

  $\begin{array}{c}L\left(T\right)=1250+{\displaystyle \underset{4}{\overset{T}{\int }}\left(-1000+500t-1500\right)dt}\\ L\left(T\right)=7250-2500T+250T^2\end{array}$

At $T=5$ , $L\left(5\right)=7250-2500\left(5\right)+250{\left(5\right)}^2=1000$ . That means, 1000 people are waiting in the line at 5 pm.

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

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

Conclusion:

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

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