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Math Help - Applied Maxima and Minima Problem

  1. #1
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    Applied Maxima and Minima Problem

    The base and top of a rectangular covered box are three times as long as they are wide. If the volume is a maximum for fixed total surface area, find the proportion of the height of the box to the short side of its base.

    Does anyone know how to solve this??
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  2. #2
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    Wink

    Let:
    x = width of box
    h = height of box
    V = Volume of box
    A = Surface Area of box
    P = Proportion of the height of the box to the side of the base
    then:
    length of box = 3x

    V = 3x^2h

    A = 2x.h+2.3x.x+2.3x.h = 8xh+6x^2=2x(4h+3x)

    P = \frac{h}{x}

    Now for constant A we can make h a function of x by re-arranging formula for A as follows:
    \frac{A}{2x}=4h+3x

    4h=\frac{A}{2x}-3x

    h=\frac{A}{8x}-\frac{3}{4}x

    And so now we can calculate V as a function of x:
    V = 3x^2h=3x^2(\frac{A}{8x}-\frac{3}{4}x) = \frac{3A}{8}x-\frac{9}{4}x^3

    And to find the maximum V we need \frac{dV}{dx}=0 so:
    \frac{dV}{dx}=\frac{3A}{8}-\frac{27}{4}x^2=0

    \frac{3A}{8}=\frac{27}{4}x^2

    A=18x^2

    But we also know that A = 2x(4h+3x) so:
    2x(4h+3x)=18x^2

    4h+3x=9x

    4h=6x

    2h=3x

    \frac{h}{x}=\frac{3}{2}

    So P=\frac{3}{2}

    Which is the solution:
    Proportion of the height of the box to the side of the base is 1.5 !
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  3. #3
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    It always helps to draw a picture. Let x be the short side and h be the height. Then the box is x by 3x by h so:

    V = 3x^2*h. This is what you want to maximize, i.e. take derivative and find critical points. But you need to use surface area to solve and substitute for h.

    Surface area is a known constant, let's call "SA". SA is the four sides plus the top and bottom. The top and bottom are 3x^2 each and there are 2 sides xh and 2 sides 3xh. So:

    SA = 3x^2 + 3x^2 + 2xh + 6xh = 6x^2 + 8xh.

    Solve for h = (SA-6x^2)/8x, substitute into V and take derivative of V with respect to x, solve for zero. That will be best x, called Xmax. Then substitute Xmax into above h equation to find Hmax and take the proportion.

    Hope this helps.

    [This question took 6 minutes to answer]
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  4. #4
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    Quote Originally Posted by Pebbles View Post
    The base and top of a rectangular covered box are three times as long as they are wide. If the volume is a maximum for fixed total surface area, find the proportion of the height of the box to the short side of its base.
    L = 3W

    V = LWH = 3W^2H

    H =  \frac{V}{3W^2}

    desired ratio is ...

    \frac{H}{W} = \frac{V}{3W^3}


    surface area ...

    A = 2(LW+WH+HL)

    A =  2\left(3W^2 + \frac{V}{3W} + \frac{V}{W}\right)

    if V is a max, then \frac{dA}{dW} = 0 ...

    \frac{dA}{dW}  = 2\left(6W - \frac{V}{3W^2} - \frac{V}{W^2}\right) = 0

    6W  = \frac{4V}{3W^2}

    W^3 = \frac{2V}{9}

    sub into the earlier expression for \frac{H}{W} ...

    \frac{H}{W}  = \frac{V}{3} \cdot \frac{9}{2V} = \frac{3}{2}
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