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Math Help - Optimization Problem

  1. #1
    Super Member
    Joined
    Dec 2008
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    509

    Optimization Problem

    Hi
    i am having trouble getting the correct answer on the following question:
    A box open at the top has a volume of 10m^3. Find the dimensions which give minimum surface area.

    (1)  V=xyw=10

    (2) A=2xy+2yw+xw

    w=\frac{10}{xy}

    sub w into (2) equation

    A=2xy+2y(\frac{10}{xy})+x({\frac{10}{xy})
    =2xy+\frac{20}{x}+\frac{10}{y}

    \frac{\partial A}{\partial x} = 2y-\frac{20}{x^2}

    \frac{\partial A}{\partial x} = 0
    (3) 2y-\frac{20}{x^2} = 0

    \frac{\partial A}{\partial y} = 2x-\frac{10}{y^2}

    \frac{\partial A}{\partial y} = 0
    (4) 2x-\frac{10}{y^2} = 0

    make x the subject of equation (4)
    2x=\frac{10}{y^2}
    x=5y^2

    sub back into equation (3)
    2y-\frac{20}{5y^2} = 0
    2y-4y^2 = 0
    2y(1-2y) = 0

    therefore

    y=0 or  \frac{1}{2}

    0 is unrealistic so don't consider it.

    x = \frac{5}{4}

    w= 16

    my answer \frac{1}{2}m , \frac{5}{4}m, 16m

    answer is cubic root 20m, cubic root 20m, 0.5 cubic root 20m

    P.S
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  2. #2
    Super Member
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    Jun 2009
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    Thanks
    4
    2x = 10/y^2, So x = 5/y^2

    2y - 20/5y^2 = 0

    2y - 4/y^2 = 0

    y^3 = 2.

    Check the simplifications.
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