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Math Help - Help with Minimum Area

  1. #1
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    Help with Minimum Area

    1) A cardboard box of 32in^3 volume with a square base and open top is to be constructed. Find the minimum area of cardboard needed.

    Since it has a square base, length and width must be the same, variable b for base.

    So volume could be written as: hb^2=32 and surface area as b^2+4bh=min.

    Once I get here, I''m not quite sure what to do.
    Any help is greatly appreciated!
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  2. #2
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    Quote Originally Posted by karisrou View Post
    1) A cardboard box of 32in^3 volume with a square base and open top is to be constructed. Find the minimum area of cardboard needed.

    Since it has a square base, length and width must be the same, variable b for base.

    So volume could be written as: hb^2=32 and surface area as b^2+4bh=min.

    Once I get here, I''m not quite sure what to do.
    Any help is greatly appreciated!
    All you considerations and calculations are OK!

    The value which should become a minimum is the area of the cardboard:

    a = b^2+4bh

    Use the side information(?) to eliminate one variable in a:

    hb^2 = 32 ~\implies~h=\dfrac{32}{b^2} . Plug in this term into a:

    a(b)=  b^2+4b\cdot \dfrac{32}{b^2} = b^2 + \dfrac{128}b

    Solve the equation a'(b) = 0 for b:

    2b - \dfrac{128}{b^2} = 0~\implies~2b^3 = 128~\implies~ b = 4

    Therefore h = 2
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