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12 January, 12:40

An airline finds that if it prices a cross-country ticket at $300, it will sell 200 tickets per day. It estimates that each $10 price reduction will result in 50 more tickets sold per day. Find the ticket price (and the number of tickets sold) that will maximize the airline's revenue

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  1. 12 January, 13:20
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    Ticket price = $170

    Number of tickets sold = 850

    Step-by-step explanation:

    Let X = the number of $10 price reduction and the number of 50 tickets increase

    Let p = price per ticket

    Let q = number of tickets sold

    Let R = revenue

    Let p (x) = price of ticket now sold

    Let q (x) = number of tickets now sold

    R (x) = p (x) * q (x)

    P (x) = 300 - 10x

    q (x) = 200 + 50x

    R (x) = (300 - 10x) (200 - 50x)

    = 60000 + 15000x - 2000x - 500x^2

    = 60000 + 13000x - 500x^2

    Differentiate with respect to x

    dR (x) / dx = 13000-1000x

    Put dR (x) / dx = 0

    0 = 13000 - 1000x

    1000x = 13000

    Divide through by 1000

    1000x/1000 = 13000/1000

    x = 13

    Therefore x = 13 for $10 price reduction and 50 tickets increase

    p (x) = 300 - 10x

    = 300 - 10*13

    = 300 - 130

    = $170

    q (x) = 200 + 50x

    = 200 + 50*13

    = 200 + 650

    = 650

    Maximum Revenue = p (x) * q (x)

    = 170 * 850

    = $144,500

    For 13 reductions, we have $170 price reduction and 850 ticket sales
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