Chapter MPS, Problem 5.11MPS

a.

To determine

To find: The quadratic function giving the softball teams profit.

Answer to Problem 5.11MPS

The quadratic function giving the softball teams profit is $P\left(n\right)=-1.5n^2+75n+375$ .

Explanation of Solution

Given information: The cost of ticket is $15 per person. For each empty seat the charge is $1.50 per person.

Calculation:

Assume the number of passengers who are not traveling in bus is n.

The revenue is $\left(60-n\right)\left(15+1.5n\right)$ .

The expense is $525$ .

The profit is,

  $\begin{array}{c}P\left(n\right)=\left(60-n\right)\left(\mathrm{15.1.5}n\right)-525\\ P\left(n\right)=-1.5n^2+75n+375\end{array}$

Thus, the quadratic function giving the softball teams profit is $P\left(n\right)=-1.5n^2+75n+375$ .

b.

To determine

To find: The minimum number of passengers required for the team not to lose money.

Answer to Problem 5.11MPS

The minimum number of passengers required for the team not to lose money is $54$ .

Explanation of Solution

Given information: The cost of ticket is $15 per person. For each empty seat the charge is $1.50 per person.

Calculation:

The profit is,

  $\begin{array}{c}P\left(n\right)=\left(60-n\right)\left(\mathrm{15.1.5}n\right)-525\\ P\left(n\right)=-1.5n^2+75n+375\end{array}$

Substitute $p\left(n\right)=0$ in above equation.

  $\begin{array}{l}0=-1.5n^2+75n+375\\ n=54.58\end{array}$

Thus, the minimum number of passengers required for the team not to lose money is $54$ .

c.

To determine

To find: The maximum profit team can earn.

Answer to Problem 5.11MPS

The maximum profit team can earn is $1312.5$ .

Explanation of Solution

Given information: The cost of ticket is $15 per person. For each empty seat the charge is $1.50 per person.

Calculation:

The profit is,

  $\begin{array}{c}P\left(n\right)=\left(60-n\right)\left(\mathrm{15.1.5}n\right)-525\\ P\left(n\right)=-1.5n^2+75n+375\end{array}$

Differentiate above equation.

  $P^{\prime }\left(n\right)=-1.5\left(2\right)n+75$

Substitute $P^{\prime }\left(n\right)=0$ in above equation.

  $\begin{array}{l}0=-1.5\left(2\right)n+75\\ n=25\end{array}$

Substitute $n=25$ in above equation.

  $\begin{array}{c}P\left(25\right)=-1.5{\left(25\right)}^2+75\left(25\right)+375\\ =1312.5\end{array}$

The required number of passenger is,

  $60-25=35$

Thus, the maximum profit team can earn is $1312.5$ .