2 variable optimization problem

**superdude** · October 18th 2009, 09:56 AM

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 $C=40xy + \frac{320}{x} + \frac{320}{y}$

the partial derivatives are $40y-\frac{320}{x^2}$ and $40x-\frac{320}{y}$ and I set them equal to 0 to get
$y=\frac{8}{x^2}$ (1) and $x=\frac{8}{y^2}$ (2)

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

**Mush** · October 18th 2009, 10:03 AM

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 $C=40xy + \frac{320}{x} + \frac{320}{y}$

the partial derivatives are $40y-\frac{320}{x^2}$ and $40x-\frac{320}{y}$ and I set them equal to 0 to get
$y=\frac{8}{x^2}$ (1) and $x=\frac{8}{y^2}$ (2)

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

Yes, you're right!

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

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

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

Square both sides:

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

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

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

**superdude** · October 18th 2009, 10:19 AM

thanks

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