Chapter 4, Problem 47RE

To determine

To find: The revenue maximize from ticket sale.

Answer to Problem 47RE

The revenue maximize from ticket sale price is $\$11.50$

Explanation of Solution

Given information:

The seating capacity of arena is $15,000$ .The ticket prize set $\$12$ ,Average attendance will increase by 1000.

Calculation:

Let X is number of attendance, then the increase in attendance is $x-11000$ ,For each increase of 1000 attendance the price decreased by $\$1$ so for additional attendance, the decrease in price will be $\frac{1}{1000}\times 1$

And the demand function is,

  $\begin{array}{c}p\left(x\right)=12-\frac{1}{1000}\left(x-11000\right)\\ =23-\frac{x}{1000}\end{array}$

The revenue function is $R\left(x\right)=xp\left(x\right)=23-\frac{x^2}{1000}$

Since $R^{\prime }\left(x\right)=23-\frac{x}{500}$ so $R^{\prime }\left(x\right)=0$

  $x=11500$

This value of X gives an absolute maximum by first derivative test.

Then ticket price is,

  $\begin{array}{c}p\left(x\right)=23-\frac{1}{1000}\left(11500\right)\\ =23-11.50\\ =11.50\end{array}$

So the price is $ $11.50$ which will maximize the revenue.