Question

A professional basketball team plays in a stadium that holds 23,000 spectators. With ticket prices at $60, the average attendance had been 18,000. When ticket prices were lowered to $55, the average attendance rose to 20,000. Based on this pattern, how should ticket prices be set to maximize ticket revenue?

Answer 1

looks like for each $5 decrease in the price, the attendance rises by 2000

let the number of $5 decreases be n
the price of a ticket = 60-5n
attendance = 18000 + 2000n

revenue = (60-5n)((18000+2000n)
factoring ....
= 5(12-n)(1000)(18 + 2n)
= 5000(12-n)(18+2n)
= 5000(216 + 6n - 2n^2)

d(revenue)/dx = 5000(6 - 4n)
= 0 for a max of revenue
4n = 6
n = 3/2
I would assume that the price change would stay as whole multiples of 5 , so I would round this up to 2

but we defined as the number of decreases, so there should be 2 decreases of $5 or a decrease of $10
The tickets should sell at $50.00

check:
at $60, attendance = 18000, revenue = 1,080,000
at $55, attendance = 20000, revenue = 1,100,000
at $50 , attendance = 22000, revenue = 1,100,000

So they actually have the same revenue at $55 as at $50
Now if we were allowed to use the actual mathematical answer of 3/2 decreases, we would have a decrease of (3/2)(5) or $7.50
for a ticket cost of 52.50 , and an attendance of
21000 and a revenue of $1,102,500

I will leave the interpretation of the question up to you.