# Thread: 2 variable optimization problem

1. ## 2 variable optimization problem[SOLVED]

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.

2. Originally Posted by 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.
Yes, you're right!

$\displaystyle y = \frac{8}{x^2} \rightarrow x = \pm \sqrt{\frac{8}{y}}$

And you also know that $\displaystyle x = \frac{8}{y^2}$

Therefore $\displaystyle \frac{8}{y^2} = \pm \sqrt{\frac{8}{y}}$

Square both sides:

$\displaystyle \frac{64}{y^4} = \frac{8}{y}$

$\displaystyle \therefore y^3 = 8 \rightarrow y = 2$

Now you can also find x by substituting this value into either equation you had at the start.

3. thanks