Question

A baseball team plays in a stadium that seats 55,000 spectators. With ticket prices at \(10, the average attendance had been 27,000. When ticket prices were lowered to \)8, the average age attendance rose to 33,000.

1. Find the demand function, assuming that it is linear.
2. How should ticket prices be set to maximize revenue?

Short answer

1. The demand function is\(y = p\left( x \right) = 19 - \frac{x}{{3000}}\).
2. The ticket price is set to maximize revenue is $9.50.

Step-by-step solution

(a) Step 1: Find the demand function

It is given that the demand function\(p\)is linear and\(p\left( {27,000} \right) = 10\),\(p\left( {33,000} \right) = 8\).

Obtain the slope as shown below:

\(\begin{aligned}{c}m = \frac{{10 - 8}}{{27,000 - 33,000}}\\ = - \frac{1}{{3000}}\end{aligned}\)

Obtain the equation of the line as shown below:

\(\begin{aligned}{c}y - 10 = \left( { - \frac{1}{{3000}}} \right)\left( {x - 27,000} \right)\\y = - \frac{x}{{3000}} + \frac{{27,000}}{{3000}} + 10\\y = - \frac{x}{{3000}} + 9 + 10\\y = - \frac{x}{{3000}} + 19\end{aligned}\)

Thus, the demand function is \(y = p\left( x \right) = 19 - \frac{x}{{3000}}\).

(b) Step 2: What ticket prices be set to maximize revenue

The revenue is shown below:

\(\begin{aligned}{c}R\left( x \right) = xp\left( x \right)\\ = x\left( {19 - \frac{x}{{3000}}} \right)\\ = 19x - \left( {\frac{{{x^2}}}{{3000}}} \right)\end{aligned}\)

Obtain the derivative of\(R\left( x \right)\)as shown below:

\(\begin{aligned}{c}R'\left( x \right) = \frac{d}{{dx}}\left( {19x - \left( {\frac{{{x^2}}}{{3000}}} \right)} \right)\\ = 19 - \left( {\frac{{2x}}{{3000}}} \right)\\ = 19 - \left( {\frac{x}{{1500}}} \right)\end{aligned}\)

When\(R'\left( x \right) = 0\)then,

\(\begin{aligned}{c}19 - \left( {\frac{x}{{1500}}} \right) = 0\\\frac{x}{{1500}} = 19\\x = 28,500\end{aligned}\)

The second derivative of the function is shown below:

\(\begin{aligned}{c}R''\left( x \right) = \frac{d}{{dx}}\left( {19 - \left( {\frac{x}{{1500}}} \right)} \right)\\ = 0 - \frac{1}{{1500}}\\ = - \frac{1}{{1500}} < 0\end{aligned}\)

As\(R''\left( x \right) < 0\)the maximum revenue happens if\(x = 28,500\). The price is shown below:

\(\begin{aligned}{c}p\left( {28,500} \right) = \left( {19 - \frac{{28,500}}{{3000}}} \right)\\ = 19 - 9.5\\ = \$ 9.50\end{aligned}\)

Thus, the ticket price is set to maximize revenue is $9.50.