Question

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

(a) Find the demand function, assuming that it is linear.

(b) How should ticket prices be set to maximize revenue?

Short answer

(a) The demand function when the attendance increases from 27000 to33000 for a $2 drop in ticket price is \(19 - 0.00033x\).

(b) The ticket price should be set at$9.5 to maximize the revenue.

Step-by-step solution

Given data

The average attendance had been 27,000 when the ticket prices were at $10. The average attendance rose to 33,000 when ticket prices were lowered to $8.

Maximization condition

The condition for minima of a function is:

\({\frac{{df\left( x \right)}}{{dx}}_{x = {x_{\max }}}} = 0\;\;\;\;\;.....\left( 1 \right)\)

(a)Step 3: Demand function

Let the attendance be x. Increase in attendance is:

\(x - 27000\)

Price drop per unit change in attendance is:

\(\frac{{10 - 8}}{{33000 - 27000}} = \frac{2}{{6000}}\)

Therefore, demand function is:

\(\begin{aligned}{c}p\left( x \right) &= 10 - \frac{2}{{6000}}\left( {x - 27000} \right)\\ &= 10 - 0.00033x + 9\\ &= 19 - 0.00033x\end{aligned}\)

The demand function is \(19 - 0.00033x\).

(b)Step 4: Maximum revenue

The revenue function is:

\(\begin{array}{c}R\left( x \right) = xp\left( x \right)\\ = 19x - 0.00033{x^2}\end{array}\)

The revenue is maximum for:

\(\begin{aligned}{c}{\frac{{dR\left( x \right)}}{{dx}}_{x = {x_{\max }}}} &= 0\\19 - 0.00066{x_{\max }} &= 0\\{x_{\max }} &= \frac{{19}}{{0.00066}}\\{x_{\max }} &\approx 28788\end{aligned}\)

The corresponding cost of ticket is:

\(\begin{aligned}{c}p\left( {28788} \right) &= 19 - 0.00033 \times 28788\\ &\approx 9.5\end{aligned}\)

The ticket price to maximize revenue is $9.5.