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Math Help - applications of differentiation

  1. #1
    Member Mr Rayon's Avatar
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    applications of differentiation

    A bushfire burns out A hectares of land, t hours after it started, according to the rule:

    A = 90t^2 - 3t^3

    After how long is the rate of change equal to 756 hectares per hour?
    __________________________________________________ _______________

    In order to answer this question I found the derivative of A = 90t^2 - 3t^3, which is dV/dt = 180t -9t^2

    And then I tried to substitute 756 to equate the derivative of A.
    180t - 9t^2 = 756
    -9t^2 + 180t -756 = 0

    But I got different answers when I tried to factorise the above. Using a calculator, I got t = -0.356466 or t = 23.5647 but...

    t = 6 or t = 14 hours...can somebody show where I went wrong? Could somebody show me the complete working out? Any help will be appreciated.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Mr Rayon View Post
    A bushfire burns out A hectares of land, t hours after it started, according to the rule:

    A = 90t^2 - 3t^3

    After how long is the rate of change equal to 756 hectares per hour?
    __________________________________________________ _______________

    In order to answer this question I found the derivative of A = 90t^2 - 3t^3, which is dV/dt = 180t -9t^2

    And then I tried to substitute 756 to equate the derivative of A.
    180t - 9t^2 = 756
    -9t^2 + 180t -756 = 0

    But I got different answers when I tried to factorise the above. Using a calculator, I got t = -0.356466 or t = 23.5647 but...

    t = 6 or t = 14 hours...can somebody show where I went wrong? Could somebody show me the complete working out? Any help will be appreciated.
    A=90t^2-3t^3

    \frac{dA}{dt}=180t-9t^2

    \frac{dA}{dt}=756

    180t-9t^2=756

    9t^2-180t+756=0

    t^2-20t+84=0

    (t-6)(t-14)=0

    t=6 hours or t=14 hours
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