# Thread: find a formula, 2 variables

1. ## 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. 2. Originally Posted by superdude 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).

3. So C=(16/(x*y))*y*x-16 ?

4. Originally Posted by aidan
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?

5. Originally Posted by superdude
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

6. 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