Question

A university is trying to determine what price to charge for tickets to football games. At a price of

​$20 per​ ticket, attendance averages

40,000 people per game. Every decrease of

​$4adds 10,000people to the average number. Every person at the game spends an average of

​$6.00 on concessions. What price per ticket should be charged in order to maximize​ revenue? How many people will attend at that​ price?

What is price per ticket?

What is the avergage attendence ?

Answer 1

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

Every decrease of $4 adds 10,000 people or

Every decrease of $1 adds 2,500 people

Cost of ticket = $20 - x

#people = 40,000 + 2,500 x

Revenue = (40,000 + 2,500 x) (20 - x) + 6 (40,000 + 2500 x)

............. = 800,000 + 10,000 x - 2,500 x² + 240,000 + 15,000 x

............. = -2,500 x² + 25,000x + 800,000

Need to maximize revenue, or the function

f(x) = -2,500 x² + 25,000x + 800,000

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

f'(x) = -5,000 x + 25,000 = 0

. . . . -5,000 x = -25,000

. . . . x = 5

f''(x) = -5,000

f''(5) = -5,000 < 0

Revenue is maximized when x = 5

Cost of ticket = $20 - x = $20 - 5 = $15

#people = 40,000 + 2,500 x = 40,000 + 2,500*5 = 40,000 + 12,500 = 52,500