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Math Help - 2 Optimization Problems

  1. #1
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    2 Optimization Problems

    An open topped rectangular box with a square base is to be constructed with a volume of 32m^{3}. Find the Dimension that require the lease amount of surface material.

    and...

    In microcomputers, most of the components are squeezed into a single box-shaped block. If the block has a length equal to twice the width and if the total surface area of the block must be 200cm^{2} in order to dissipate the heat produced, find the dimensions for the maximum volume of the block.

    I couldn't figure out how to get equations out of these two problems.


    Thanks.
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  2. #2
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    Quote Originally Posted by johntuan View Post
    ...

    In microcomputers, most of the components are squeezed into a single box-shaped block. If the block has a length equal to twice the width and if the total surface area of the block must be 200cm^{2} in order to dissipate the heat produced, find the dimensions for the maximum volume of the block.

    ...
    Hello,

    surface area of a rectangular block: a = 2\cdot l\cdot w + 2\cdot l\cdot h + 2\cdot w\cdot h

    Since l = 2w and a = 200 you get:

    200 = 2\cdot 2w \cdot w + 2\cdot 2w \cdot h + 2\cdot w\cdot h~\implies~200 = 4w^2 + 6wh~\implies~h = \frac{200}{6w} - \frac{4w^2}{6w}

    The volume of a rectangular block is V=l \cdot w \cdot h

    Plug in the values you already know:

    V(h)=2w^2 \cdot \left( \frac{200}{6w} - \frac{4w^2}{6w} \right)

    Expand the bracket and calculate the minimum using the first derivative ....and so on.
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  3. #3
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    How would you expand the bracket?
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  4. #4
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    Quote Originally Posted by johntuan View Post
    How would you expand the bracket?
    Hello,

    multiply and cancel equal factors:

    <br />
V(h)=2w^2 \cdot \left( \frac{200}{6w} - \frac{4w^2}{6w} \right) = \frac{200}{3} w - \frac43 w^3......... Therefore

    V'(h) = \frac{200}{3}-4w^2........ To calculate the minimum V'(h) = 0:

    w = \sqrt{\frac{50}{3}}\approx 4.08... ........ Now plug in this value to calculate l and h.

    \left(l=\sqrt{\frac{200}{3}},~h=\frac{20}{9} \cdot \sqrt{6} \right)
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