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Math Help - Optimization

  1. #1
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    Optimization

    A rectangle storage container has a volume of 10 meters cubed and no top. The length is twice the width, the base material costs $10 per meter cubed and the side material costs $6 per meter cubed. Calculate the dimensions of the box that minimize total cost of materials.
    <br />
V = lwh = 10<br />
    2l = w
    V = 2l^2h

    SA = (10)(2l^2) + 4[6(lh)]
    SA = 20l^2 + 24lh
    <br />
H = 5/l^2<br />

    SA = 20l^2 + 120/l
    SA' = 40l - 120/l^2 = 0
    l = cubed root of 3?

    is this much correct?
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  2. #2
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    Quote Originally Posted by millerst View Post
    A rectangle storage container has a volume of 10 meters cubed and no top. The length is twice the width, the base material costs $10 per meter cubed and the side material costs $6 per meter cubed. Calculate the dimensions of the box that minimize total cost of materials.
    V = LWH

    L = 2W

    V = 2W^2H = 10

    H = \frac{5}{W^2}

    C = 2W^2(10) + 2WH(6) + 4WH(6)<br />

    C = 20W^2 + 36WH

    C = 20W^2 + \frac{180}{W}

    \frac{dC}{dW} = 40W - \frac{180}{W^2} = 0

    W^3 = \frac{9}{2}

    W = \sqrt[3]{\frac{9}{2}}
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