Question

A university is trying to determine what price to charge for tickets to football games. At a price of ​$22 per​ ticket, attendance averages 40,000 people per game. Every decrease of ​$2 adds 10,000 people to the average number. Every person at the game spends an average of ​$3.00 on concessions. What price per ticket should be charged in order to maximize​ revenue? How many people will attend at that​ price?

Answer 1

SOLUTION:

Every fall of $2 adds 10,000 people or Every fall of $1 adds 5,000 people

Cost of ticket = $22 - x

Number of people = 40,000 + 5,000x

Revenue = Number of people * cost of ticket + Number of people * $6

f(x) = (40,000 + 5,000x) ($22-x) + $3 (40,000 + 5,000x)

f(x) = 880,000 - 40,000x + 110,000x - 5000x^2 + 120,000 + 15,000x

f(x) = - 5000x^2 + 85,000x + 1,000,000

We know that f(x) is maximized at x when f'(x) = 0

f'(x) = -10,000x + 85,000 = 0

x = 85,000/10,000 = 8.5

Revenue would be maximized when x = 8.5

Cost of ticket = $22 - x = $22 - $8.5 = $13.5

Number of people = 40,000 + 5,000x = 40,000 + 5,000 * 13.5 = 107,500