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Math Help - Find dimensions to minimize total cost of materials

  1. #1
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    Find dimensions to minimize total cost of materials

    Problem: A rectangular box is to be open-topped with a volume of 600 in^2. The material for its bottom costs $ 6/in^2, and the material for its four sides costs $ 5/in^2. Find the dimensions of the box that minimize the total cost of the material needed to construct the box.

    My Solution:
    Let h = height, l = length, w = width. Remember that the volume of a rectangular prism is V = lwh. We can solve for h = \dfrac{V}{lw} = \dfrac{600}{lw}.

    Now we can write our equation for the total cost as:
    C = ($ 6/in^2)lw + ($ 5/in^2)2wh + ($ 5/in^2)2lh

    = 6lw + 10wh + 10lh

    = 6lw + 10w\dfrac{600}{lw} + 10l\dfrac{600}{lw}

    = 6lw + \dfrac{6000}{l} + \dfrac{6000}{w}. (1)

    We can take the derivative with respect to l:
    \dfrac{\partial C}{\partial l} = 6w - \dfrac{6000}{l^2} + 0 = 0.

    so 6w = \dfrac{6000}{l^2} which gives us w = \dfrac{1000}{l^2}.

    so we can substitute w back into (1) and we have:
    = 6l\dfrac{1000}{l^2} + \dfrac{6000}{l} + \dfrac{6000}{\dfrac{1000}{l^2}}

    = \dfrac{6000}{l} + \dfrac{6000}{l} + 6l^2

    = \dfrac{12000}{l} + 6l^2.

    Once again we can take the derivative with respect to l:
    \dfrac{\partial C}{\partial l} = \dfrac{-12000}{l^2} + 12l = 0.

    so 12l = \dfrac{12000}{l^2} which gives us l^3 = 1000 and thus l = +10 or -10, however l must be positive so l=10.

    Therefore we have: l=10, w = \dfrac{1000}{l^2} = \dfrac{1000}{100} = 10, and h = \dfrac{600}{lw} = \dfrac{600}{100} = 6.

    Is that correct? Does that make sense?
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  2. #2
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    Re: Find dimensions to minimize total cost of materials

    Hey vidomagru.

    That is spot on: well done .

    Intuitively your answer also makes sense since a square will provide the biggest area and/or volume under optimization if it has to be rectangular shape.
    Thanks from vidomagru
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