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Math Help - optimization question

  1. #1
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    optimization question

    This one has been kicking my rear for too long so any help is appreciated.


    A box (top included) is to be made to hold exactly 216 cubic inches. It is to have a square base. Find the dimensions of the box that has minimum surface area.
    Volume= lwh
    Surface Area=2lw + 2lh + 2wh
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Exiab View Post
    This one has been kicking my rear for too long so any help is appreciated.


    A box (top included) is to be made to hold exactly 216 cubic inches. It is to have a square base. Find the dimensions of the box that has minimum surface area.
    Volume= lwh
    Surface Area=2lw + 2lh + 2wh
    dont make it too hard...all sides are x..sicne in a box all sides are equal since it is a cube...and remember that there is not top
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  3. #3
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    dont make it too hard...all sides are x..sicne in a box all sides are equal since it is a cube...and remember that there is not top
    Actually, a box can be any rectangular shape. This problem states that it has a square base meaning that the length and width are equal. Let's call the length or width a side of the square base, s.

    V=hs^2
    SA=2s^2+2sh+2hs \rightarrow 2s^2 + 4sh

    The volume is 216 inches cubed...

    hs^2=216

    You want to minimize the surface area. Key word here is minimize, so you must derive the surface area equation with respect to one of the sides.

    Let's solve the surface area equation for one variable, say s.

    Rearranging our prior equation for volume: h = \frac{216}{s^2}

    Plug that in for h in our SA equation.

    SA=2s^2+\frac{216*4}{s}

    Differentiate this with respect to s and set it equal to zero. Solve for s. Whichever value of s gives the minimum SA (plug s back into original SA equation) is your answer.
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