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Math Help - minimising volume

  1. #1
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    Question minimising volume

    A company wishes to manufacture a box with a volume of 32 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material.
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  2. #2
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    Quote Originally Posted by Matho View Post
    A company wishes to manufacture a box with a volume of 32 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material.
    V = L \cdot W \cdot H

    L = 2W

    32 = 2W \cdot W \cdot H

    \frac{16}{W^2} = H

    surface area ...

    A = (2W \cdot W) + 2(W \cdot H) + 2(2W \cdot H)

    get A in terms of a single variable ( W), find \frac{dA}{dW}, set it equal to 0, and find the value of W that minimizes the area.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    V = L \cdot W \cdot H

    L = 2W

    32 = 2W \cdot W \cdot H

    \frac{16}{W^2} = H

    surface area ...

    A = (2W \cdot W) + 2(W \cdot H) + 2(2W \cdot H)

    get A in terms of a single variable ( W), find \frac{dA}{dW}, set it equal to 0, and find the value of W that minimizes the area.

    Thanks for replying me.
    I got da/dW' = 4w - 96/w^2 but how do I get equal 0?
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  4. #4
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    4W - \frac{96}{W^2} = 0

    4W = \frac{96}{W^2}

    W^3 = 24

    W = 2\sqrt[3]{3}
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