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March 27, 2017

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can anyone please help me set this problem up, I can do the math but I have tried for hours...
Air fare club offers membership at $300.00, at least 50 people must join.For every member over 50 their fare will be reduced by $2 for every member. Due to space limits only 125 passengers will be allowed. how many memberships will maximize the revenue?

  • algebra - ,

    current rate = $300
    number of members = 50

    let the number of additonal member be n

    so actual number of members = n+50
    cost per member = 300 - 2n

    Revenue = number of members x membership fee
    = (n+50)(300 - 2n)
    = 300n - 2n^2 + 15000- 100n
    = -2n^2 + 200n + 15000
    = -2(n^2 - 100n + 2500 - 2500 ) + 15000
    = -2( (n-50)^2 - 2500) + 15000
    = -2(n-50)^2 + 5000 + 15000
    = -2(n-50)^2 + 20000

    This is a parabola opening downward with a vertex of (50,20000)

    So the maximum of Revenue is $20,000 when n = 50
    So 50 addition to the original 50 should join.

    The maximum revenue is obtained when 100 members are in.

    check:
    at 100 members cost is 300 - 2(50) = 200
    revenue is 100(200) = 20000
    at 99 members cost per member is 300-2(49) = 202
    revenue is 99(202) = 19998 which is less than 20000
    at 101 members cost per member is 300-2(51) = 198
    revenue is 101(198) = 19998 , which is again less than 20000
    My answer is correct.

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