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29 December, 13:56

A movie theater estimates that for each $0.50 increase in ticket price, the number of tickets sold decreases by 60. The current ticket price of $16.50 yields 1800 tickets sold. Set up an equation to represent the daily revenue. What price should the theatre charge for tickets in order to maximize daily revenue

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  1. 29 December, 17:24
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    The daily revenue equation is 29700 - 90x - 30x²

    The price that should be charged to maximize daily revenue is $15.75

    Explanation:

    Revenue is a function of price multiplied by the quantity. Thus, the equation for revenue can be written as,

    Let x be the number of times there is an increase in price of $0.5.

    The price function is = 16.5 + 0.5x

    The demand function is = 1800 - 60x

    Revenue = (16.5 + 0.5x) * (1800 - 60x)

    Revenue = 29700 - 990x + 900x - 30x²

    Revenue = 29700 - 90x - 30x²

    To calculate the price that will yield maximum revenue, we need to take the derivative of this equation of revenue.

    d/dx = 0 - 1 * 90x° - 2 * 30x

    0 = - 90 - 60x

    90 = - 60x

    90 / - 60 = x

    x = - 1.5

    The price needed to maximize revenue is,

    p = 16.5 + 0.5 * (-1.5)

    p = 16.5 - 0.75

    p = 15.75

    The demand at this price is = 1800 - 60 * (-1.5)

    Demand = 1800 + 90 = 1890
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Home » Business » A movie theater estimates that for each $0.50 increase in ticket price, the number of tickets sold decreases by 60. The current ticket price of $16.50 yields 1800 tickets sold. Set up an equation to represent the daily revenue.
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