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Math Help - mamimize rectangle area

  1. #1
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    mamimize rectangle area

    An open box is made from a rectangular piece of cardboard by cutting equal squares from each corner and turning up the sides. Write the volume of the box as a function of x if the material is 20 inches by 12 inches. What is maximum volume of the box?

    This was one problem that stumped me alot.
    Thank you!!
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  2. #2
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    x appears to be the length of the sides of the removed squares.
    When the side lengths are cut out, the piece of cardboard looks like this and is then folded along the dotted lines to create an open topped box.

    mamimize rectangle area-net.jpg
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  3. #3
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    so what is the the volume of the box as a function of x and the maximum volume of the box?

    thankss
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  4. #4
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    The volume of the box is length \times width \times height
    The maximum volume of the box will be at either a stationary point or at an endpoint.
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  5. #5
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    After we cut the squares of length x, the base of the box will be a rectangle with sides 20-2x and 12-2x (because we've taken a square off at both ends of each side). The height of the box will simply be x. So the volume as a function of x would be V(x)=x(20-2x)(12-2x).

    To find the max volume, ie. optimize V, differentiate and set the derivative to zero:

    V'(x)=(20-2x)(12-2x)-2x(12-2x)-2x(20-2x)=0

    (10-x)(6-x)-x(6-x)-x(10-x)=0 (divide by 4 everywhere)

    60-16x+x^2-6x+x^2-10x+x^2=0

    3x^2-32x+60=0

    Solve this using the quadratic equation and you'll get your answer (don't forget to throw away the negative answer).
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  6. #6
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    Strictly speaking, you still need to confirm that the solution found by ss103's method is a maximum rather than a minimum or stationary point of inflection. A good way to do this is checking that the function is smaller at the end points ie. when x=0 and 6 as these are the minimum and maximum lengths of the cuts.
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