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Math Help - Revenue

  1. #1
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    Revenue

    The revenue of McMenamy's fish shanty is approximately

    R(t)=2(5-4cos (pi/6)t) (0<t<12)

    during the t week (t=1 corresponds to the first week of June) where R is measured in thousands of dollars. On what week is the marginal revenue zero?
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  2. #2
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    Quote Originally Posted by lemontea View Post
    The revenue of McMenamy's fish shanty is approximately

    R(t)=2(5-4cos (pi/6)t) (0<t<12)

    during the t week (t=1 corresponds to the first week of June) where R is measured in thousands of dollars. On what week is the marginal revenue zero?
    Marginal R is dR/dt = R'(t)

    R(t) = 2{5 -4cos[(pi/6)t]}
    R'(t) = 2[(-4)(-sin[pi/6)t] *(pi/6)]
    R'(t) = (8pi/6)sin[(pi/6)t]

    When R'(t) = 0.
    0 = (8pi/6)sin[(pi/6)t]
    0 = sin[(pi/6)t]
    (pi/6)t = arcsin(0)
    (pi/6)t = 0, or 2pi

    When (pi/6)t = 0,
    t = 0 -----------------**

    When (pi/6)t = 2pi,
    t = (2pi)/(pi/6) = 12 ----**

    Therefore, the marginal revenue is zero at week zero and week 12. ----answer.
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  3. #3
    MHF Contributor kalagota's Avatar
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    marginal revenue is the derivative of the revenue function right?

    so, solve t such that R'(t)=0.

    EDIT: didn't ticbol's solution..

    in addition, t=6 is also a solution since arcsin 0 = 0, pi, or 2pi..

    EDIT: i mean, "didn't see ticbol's solution.."
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  4. #4
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    Quote Originally Posted by kalagota View Post
    marginal revenue is the derivative of the revenue function right?

    so, solve t such that R'(t)=0.

    EDIT: didn't ticbol's solution..

    in addition, t=6 is also a solution since arcsin 0 = 0, pi, or 2pi..
    Yes, I forgot about arcsin(0) = pi also.

    Thanks.
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