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7 November, 17:15

1. A gym has 200 members who pay $30 per month for unlimited use of the gym's equipment. A survey of the members indicates that for each $5 increase in the monthly fee, the gym will lose 20 members. This means that the revenue R from fees, which is currently $6000 per month, will become, where f is a whole number of $5 fee increases. Write and solve a quadratic inequality to answer the questions: For what numbers of $5 fee increases will the revenue from fees actually be less than its current value?

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  1. 7 November, 18:57
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    200 members pay $30 each

    Current revenue = 200 x $30

    = $6000

    For each $5 increase in fee, 20 members will leave

    For i number of $5 increase, 20i members will leave

    New number of members = 200 - 20i

    For i number of $5 increase, new fee = 30 + 5i

    Therefore, new revenue = (200 - 20i) x (30 + 5i)

    = 200 (30 + 5i) - 20f (30 + 5i)

    = 6000 + 1000i - 600i - 100i²

    = 6000 + 400i - 100i²

    In the form of a quadratic equation,

    -100i² + 400i + 6000

    R (i) = - 100i² + 400i + 6000

    This is not our answer yet.

    We are told to write a quadratic inequality for what numbers of $5 fee increases will the revenue from fees actually be less than its current value

    Current value = 6000

    Therefore, we are looking for i, when R (i) < 6000

    Remember, R (i) = - 100i² + 400i + 6000

    Therefore,

    R (i) < 6000 = - 100i² + 400i + 6000 < 6000

    We are to solve the quadratic inequality,

    -100i² + 400i + 6000 < 6000

    ⇒ - 100i² + 400i < 0 (Subtract 6000 from both sides)

    ⇒ 400i < 100i² (Add 100i to both sides)

    ⇒ 4i < i² (Divide both sides by 100)

    ⇒ 4 < i (Divide both sides by i)

    4 4

    Therefore, for more than 4 numbers of $5 increase in the fees, the revenue from fees will actually be less than its current value.
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