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Thread: Optimization Problem

  1. #1
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    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 $\displaystyle 10m^3$. Find the dimensions which give minimum surface area.

    (1)$\displaystyle V=xyw=10$

    (2) $\displaystyle A=2xy+2yw+xw$

    $\displaystyle w=\frac{10}{xy}$

    sub w into (2) equation

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

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

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

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

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

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

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

    therefore

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

    0 is unrealistic so don't consider it.

    $\displaystyle x = \frac{5}{4}$

    $\displaystyle w= 16$

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