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Math Help - Need help with some question

  1. #1
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    Need help with some question

    A company has found that if it produces x thousand units/day of a product, The total cost would be T(x) thousand $, where T(x) = x^3 - 6x^3 + 13x + 15, when 0<x<9. They sell it for 28 $/unit.

    a) State the profit V(x) thousand $ as a function of x.
    b) Find the maximum profit per day.
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  2. #2
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    I'm assuming you meant to type T(x)=x^3-6x^2+13x+15
    So profit is sales-cost. Therefore, V(x)=28000x-(x^3-6x^2+13x+15)=-x^3+6x^2+27987x-15

    To find the maximum, you must first differentiate giving you V'(x)=-3x^2+12x+27987.
    Then find the critical numbers by setting V'(x) to 0.

    0=x^2-4x-9329
    Quadratic formula gives you the critical values of x=98 and x=-94 but those aren't in the interval (0,9). So the maximum is when x is either 0 or 9. You can eliminate 0 because they would obviously make no profit if they didn't make a product. So plug x=9 into the profit equation.

    V(9)=-9^3+6*9^2+27987*9-15=-729+486+251883-15=251625
    The maximum profit is $251625
    Last edited by Keithfert488; November 3rd 2009 at 07:02 PM. Reason: revising
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  3. #3
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    The answer to a) was -x^3 -6x^2 +15x -15. Thats because they used 28x, instead of 28000x.

    After that i couldn't get the answer to b), they get 85 000. I dont really understand how.
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  4. #4
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    So V(x)=-x^3-6x^2+15x-15

    First you need to find the critical numbers and see if they are in your interval (0,9). To find the critical numbers, derive the equation and set it equal to 0.

    V'(x)=-3x^2-12x+15
    0=x^2+4x-5=(x+5)(x-1)
    x=1 x=-5

    So you have a critical value at x=1. To find the maximum, you must plug in the endpoints of your interval (i.e. x=0 and x=9) and the critical number within the interval (i.e. x=1) into the original equation and find which one is the highest. That is your maximum.

    V(0)=-15
    V(1)=-1-6+15-15=-7
    V(9)=-9^3+6*9^2+15*9-15=-729+486+135-15=-123

    Well, that's what I got and it doesn't make sense. Can anyone else help? I am pretty sure that I have the right method of getting the maximum. I just don't understand how the answer could be negative.
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  5. #5
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    I have figured out where i was wrong! Thanks anyways!
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  6. #6
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    Thank you very much!

    Thank you very much!
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