What ticket price would maximize revenue

Question

A baseball team plays in a stadium that holds 56000 spectators. With the ticket price at $12 the average attendance was 24000. When the price dropped to $9, the average attendance rose to 28000. Assume that attendance is linearly related to ticket price. What ticket price would maximize revenue?

Answer 1

Let A be attendance, T be ticket price and R be revenue.

Your Revenue, R = AT which is what you want to maximize.

You are given two points and that the relationship is linear. There are a variety of ways to find the line where A is the dependent variable and T is the independent variable:

Slope is constant for all points on line (A-28K)/(T-9) = (24K-28K)/(12-9) (K just means 1000)

or A - 28K = (-4K/3)(T-9) and A = (-4K/3)T + 40K

The function you want to maximize is R = AT =((-4K/3)T + 40K)T

Usually, one would use calculus to find the T for max R, but the key words indicate that you are in pre-calculus. Since this is a quadratic that is curved downwards, the vertex is the maximum value. You can complete the square to put this in vertex notation:

Let's distribute and multiply in order to make the square term -T²

R = -4K/3*(T² - 30T)

In order to complete the square we need to add (30/2)² inside the parentheses and subtract what we added in order to not change the equation:

R = -4K/3*(T² - 30T + 225) + 4K*225/3

R = -4K/3*(T - 15)² + 4K*225/3 (The vertex is at T = 15) and the R(T=15) = (4K/3)15² = 300K which you get from plugging T=15 into the original equation or just taking the far right term in this equation.

Hope that helped. Take care.