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Math Help - find a formula, 2 variables

  1. #1
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    find a formula, 2 variables [SOLVED]

    There's a box with 16 cubic meter volume. Height z, depth y, width x. The sides cost $10/sq meter and top and bottom cost $20/sq meter. Write an equation for the cost (C) of the box using x and y.

    I'm bad at coming up with formulas, is there a general procedure or any places where I should start?

    I've got
    sides=2*(z*y)*$10
    front/back=2*(x*z)*10
    top/bottom=2*(y*x)*20
    and I think I should add them so cost=sides+front/back+top/bottom
    This won't work so well as I have no clue how to incorporate the volume and I'm not allowed to use z.
    Last edited by superdude; September 28th 2009 at 04:16 PM. Reason: solved
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  2. #2
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    Quote Originally Posted by superdude View Post
    There's a box with 16 cubic meter volume. Height z, depth y, width x. The sides cost $10/sq meter and top and bottom cost $20/sq meter. Write an equation for the cost (C) of the box using x and y.

    I'm bad at coming up with formulas, is there a general procedure or any places where I should start?

    I've got
    sides=2*(z*y)*$10
    front/back=2*(x*z)*10
    top/bottom=2*(y*x)*20
    and I think I should add them so cost=sides+front/back+top/bottom
    This won't work so well as I have no clue how to incorporate the volume and I'm not allowed to use z.
    z*y*x = 16
    Since the volume is fixed, for any two values you choose, the third value can be determined.
    z = 16/(x*y)

    In your equations, any where you have a z, replace it with 16/(x*y).
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  3. #3
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    So C=(16/(x*y))*y*x-16 ?
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  4. #4
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    Quote Originally Posted by aidan View Post
    z*y*x = 16
    Since the volume is fixed, for any two values you choose, the third value can be determined.
    z = 16/(x*y)

    In your equations, any where you have a z, replace it with 16/(x*y).
    what do you mean "In your equation" because if I do 16=\frac{16}{xy}\times x\times y that simplifies to 16=16?
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  5. #5
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    Quote Originally Posted by superdude View Post
    what do you mean "In your equation" because if I do 16=\frac{16}{xy}\times x\times y that simplifies to 16=16?
    sub it into your cost equation
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  6. #6
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    ok so

    \begin{aligned}C(x,y) &=2(yx)(20)+2(yz(10))+2(xz)(10) \\ & = 40yx+2(y(\frac{16}{yx}))10+2(x(\frac{16}{yx}))10\\ &=40yx+\frac{320}{x}+\frac{320}{y}\end{aligned}

    The next thing I need to do is find the minimum cost. So I take f_x and f_y to get 40-\frac{320}{x^2} and 40x-\frac{320}{y^2} respectively. Here I get confused because if I equate them with 0 I still can't solve for y (or x) because x (or y) is a constant but still an unknown in the equation.

    Found it out. Thanks everyone
    Last edited by superdude; September 28th 2009 at 04:14 PM. Reason: fixed LaTeX
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