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Math Help - optimization for a box

  1. #1
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    optimization for a box

    Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by Link88 View Post
    Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.
    Let the base be m and height be n. I am assuming the problem indicates square base is used.

    V = mmn = m^2n

    Assuming the box has an open top, the surface area is

    m^2 + 4mn = 64

    Solve for n from the surface area, then plug the expressions into the volume formula.

    n = \frac{64-m^2}{4m}

    V = m^2 \frac{64-m^2}{4m} = \frac{64m-m^3}{4}

    Find the first derivative of volume:
    V' = 8 - 0.75 m^2 = 0<br />
    Then solve for m.

    The story changes if this is a closed top box.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Link88 View Post
    Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.
    So the volume of a box is

    V=lwh

    Subject to the surface area

    S=2lw+2lh +2wh=64

    F(l,w,h,\lambda) =lhw-\lambda(2lw+2lh+2wh-64)

     \nabla F=0

    hw-\lambda(2w+2h)=0

    lh-\lambda(2l+2h)=0

    lw-\lambda(2l+2w)=0

    0=(2lw+2lh+2wh-64)

    Now multiply the first equation by l and the 2nd by -w and add them together

    lwh-\lambda(2lw+2lw)=0
    -lwh+\lambda(2lw+2wh)=0
    -2lw+2wh=0 \iff 2w(-l+h)=0 \iff l=h

    Doing something similar you can show that w=l=h

    Plugging into the last equation we get.

    2l^2+2l^2+2l^2-64=0 \iff 6l^2=64 \iff l^2 =\frac{32}{3} \iff l=4 \sqrt{\frac{2}{3}}

    So the maximal volume will be a cube.

    I hope this helps
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  4. #4
    Senior Member apcalculus's Avatar
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    Nice going, TheEmptySet... I was way off here. A calculus subforum for single-variable people like me? Just kidding.
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