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**superdude** Find the minimum cost of a box (with dimensions xyz) that holds 16 m^3. The top and bottom cost $20 and the sides cost $10.

So the formula for the cost is $\displaystyle C=40xy + \frac{320}{x} + \frac{320}{y}$

the partial derivatives are $\displaystyle 40y-\frac{320}{x^2}$ and $\displaystyle 40x-\frac{320}{y}$ and I set them equal to 0 to get

$\displaystyle y=\frac{8}{x^2}$ (1) and $\displaystyle x=\frac{8}{y^2}$ (2)

Then I do something like (1)+(2) to get $\displaystyle x^3=8$ but I get lost along the way.