Question

Maximizing revenue. Edwards University wants to determine what price to charge for tickets to football games. At a price of 518 per ticket, attendance averages 40,000 people per game. Every decrease of 53 to the ticket price adds 10,000 people to the average attendance. Every person at a game spends an average of 54.50 on concessions.

What price per ticket should be charged to maximize revenue?
How many people will attend at that price?

Answer 1

Price per ticket to maximize revenue = $337.8

Total number of people at that price = 74000 people

Explanation:

Let x represent the number of people as a result of decrease of 53 to the ticket price. Therefore, the price of a single ticket is;

M(x) = 518 - 53x

Similarly, we can say number of visitors of the game is;

T(x) = 40000 + 10000x

We are told that every person at a game spends an average of 54.50 on concessions. Therefore,

Total revenue will be;

R(x) = (M(x) × T(x)) + 54.5(T(x))

R(x) = ((518 - 53x) × (40000 + 10000x)) + 54.5(40000 + 10000x)

R(x) = (20720000 - 2120000x + 5180000x - 530000x²) + 2180000 + 545000x

R(x) = 20720000 + 3060000x - 530000x² + 2180000 + 545000x

R(x) = -530000x² + 3605000x + 22900000

To find the value of x for which the price is maximum, let's find the derivative of R(x) and equate to zero

dR/dx = -1060000x + 3605000

At dR/dx = 0,we have;

-1060000x + 3605000 = 0

1060000x = 3605000

x = 3605000/1060000

x = $3.4

Therefore price per ticket for the revenue to be maximum is;

M(3.4) = 518 - 53(3.4) = $337.8

Total attendance at that price will be;

T(3.4) = 40000 + 10000(3.4)

T(3.4) = 40000 + 34000

T(3.4) = 74000