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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?

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  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.

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