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Math Help - optimization of a box

  1. #1
    Junior Member surffan's Avatar
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    optimization of a box

    the length of a cedar chest is twice its width. the cost/dm^2 of the lid is four times the cost dm^2 of the rest of the cedar chest. if the volume of the chest is 1440 dm^3 find the dimensions so that the cost is a minimum.

    this is what i got so far

    V=L x W x H
    1440=2x(x)h
    1440=2x^2(h)
    h=720/x^2

    C=4c(area of lid) + c(area of base) + c(area of sides)
    optimization of a box-equations.bmp


    That is what i got for cost and for the derivative, i'm having trouble solving for x and i should be able to get the dimensions once i have the value of x
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  2. #2
    Senior Member
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    You're well on your way to the solution.

    You have too many constants in your derivative of the cost function. There should only be 1 2880 and 1 1440. Otherwise it seems right.

    Set C' = 0 and solve for x. The 'c' cancels out (you could also have set it equal to 1 at the start to make life easier - your answer doesn't depend on the actual cost of the wood, just the relative cost).
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