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Math Help - Optimization problem

  1. #1
    MHF Contributor alexmahone's Avatar
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    Optimization problem

    Three large squares of tin, each with edges 1m long, have four small, equal squares cut from their corners. All twelve resulting small squares are to be of the same size. The three large cross-shaped pieces are then folded and welded to make boxes with no tops, and the twelve small squares are used to make two small cubes. How should this be done to maximize the total volume of all five boxes?

    ------------------------------------------------------------

    My working:

    V=2x^3+3(1-2x)^2x

    V=2x^3+3(1+4x^2-4x)x

    V=2x^3+3x+12x^3-12x^2

    V=14x^3-12x^2+3x

    \frac{dV}{dx}=0 for maxima.

    42x^2-24x+3=0

    14x^2-8x+1=0

    x=0.3867 or x=0.1846

    ------------------------------------------------------------

    However, the answer given in the book is:

    0.25 m^3 (all cubes, no open topped boxes)

    Please help!
    Last edited by alexmahone; July 21st 2010 at 11:18 PM.
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  2. #2
    Senior Member
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    When finding an absolute maximum value over an interval, you must consider the points where the derivative equals zero (which you did) and ALSO the endpoints of the interval. In this case, you must consider x=0 and x=0.5 when trying to determine the maximum value because these are the smallest and largest allowable values for x.
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