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Math Help - Optimization of a box

  1. #1
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    Optimization of a box

    The length of a cedar chest is twice its width. the cost of the lid is four times the cost of the rest of the cedar chest. If the volume is 1440 dm^3, find the dimensions of the box so that the cost is a minimum.


    V=LxWxH
    1440=2x(x)H

    H=2x^2/1440


    Solve for total Cost:

    _____[Lid]____[Base]___[Long Sides]____[Short Sides]___

    c=4c(2x^2)+c(2x^2)+c(2x)(2x^2)/1440+c(x)(2x^2)/1440


    Find the derivative:

    c'=16cx+4cx-8cx(1440^-2)-4cx(1440^-2)

    Let c' = 0 to solve for x:

    This again is where it get lost, letting the function = 0 is an ongoing problem for me.
    Last edited by Butum; April 19th 2010 at 04:34 AM.
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  2. #2
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    I went over the question again on lunch break, I realized I made some errors.

    Solve for h:
    v=lxwxh
    1440cm3=2x(x)+h
    h=1440cm3/(2x^2 )

    find the cost.
    c=Lid+Base+Long sides+Short sides
    c=4c(2x^2 )+c(2x^2 )+c 1440/(2x(2x^2))+c 1440/(x(2x^2))

    Find the derivative:
    c^'=16cx+4cx+c(-1)(1440)(2x) (2x^2 )^(-2)+c(-1)(1440)(x) (2x^2 )^(-2)

    Let c' = 0 and solve for x.

    0=16cx+4cx+c(-1)(1440)(2x) (2x^2 )^(-2)+c(-1)(1440)(x) (2x^2 )^(-2)

    I ended up with:
    x=5sqrt6460
    x= 401.87
    I don't feel like that's right. Once I get the value for x I should be good to solve for the dimensions.
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  3. #3
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    It's very hard to determine what you are doing! You should always start by defining your variables.

    You have "c" on the left of the equation meaning, I guess, "total cost" but then you also have "c" on the right meaning, I think, the cost of each part. But also you say "the bottom cost twice the rest of the box". Do you mean "per square meter"? Otherwise you can't minimize the cost by changing the size.

    Just take the "c" on the right to be "1" since its specific value is not important.

    I presume that "x" is the width so that the length is 2x and the height is H= 1440/2x^2= 720/x^2.

    cost of lid= 4(2x^2)

    cost of bottom= (2x^2)

    cost of the long sides is NOT 1440/(2x(2x^2)). For one thing, you have all the measurements in the denominator. Making the box larger does NOT make the area of a side less or make the side less expensive! Also, since the 1440 is in "length cubed" that reduces to "1/length", not an area measurement. The area of the long side is the length, 2x, multiplied by the height, 1440/(2x^2) and so is (2)(1440/(2x))= 1440/x (which has units of "length squared". Since there are two long sides, that gives 2880/x as the cost of the two long sides.

    Similarly, the cost of the two short sides is 2x(1440/2x^2)= 1440/x.

    The total cost is c= 8x^2+ 2x^2+ 2880/x+ 1440/x= 10x^2+ 4320/x.

    Differentiate that and set equal to 0.
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