# Minimizing cost

The math problem is: A large bin for holding heavy material must be in the shape of a box with an open top and a square base. The base will cost 9 dollars a square foot and the sides will cost 11 dollars a square foot. If the volume must be 120 cubic feet. Find the dimensions that will minimize the cost of the box's construction.

I set up the equations the way my teacher taught me and they look like this:

$x=\$9$,$h=\$11$, $V=120ft^3$, $V=x^2h$, $SA= x^2+4xh$, $C= 81x^2+44xh$

Then I try to solve the equations by solving for one of the variables (I believe it's called 'solving them simultaneously'):

$120=x^2h$
$120/(x^2)=h$

Then I plug that into the other equation:

$C=81x^2+44x\times120/(x^2)$

$C=81x^2+5280x^-1$

$162x-5280x^-2$

$162x-5280/(x^2)=0$

Here is where I start to have trouble. I solve for $x$, plug it back into the equation $120/(x^2)=h$ and get an answer, but when I plug them into the answer box, they come up as being wrong (I end up with $x=3.19428$ and $h=37.567$). What do I do after getting $162x-5280/(x^2)=0?$ Should I move the $-5280/(x^2)$ to the other side of the equation? Because I'm not sure what to do with the "$x$'s" after that is done. I've tried multiplying either side by them and dividing either side by them, but I get $x^3$=__ and it never seems to be the right answer.

Thanks

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Uhh... 11 is the cost/sq foot of the side material; h is the height – DJohnM Feb 22 '13 at 2:31
You've done this almost completely correctly. Your only error is the cost function: You gave the cost of the base as 81x^2, but the cost is only 9 dollars per square foot, so it should be 9x^2 instead. – Alex Zorn Feb 22 '13 at 3:13

Set up two equations: $$V = x^2 h=120$$and $$Cost=9(x^2)+11(4xh)$$Rearrange the first to give $$h=120/x^2$$ and substitute in the Cost equation$$Cost=9x^2+5280/x$$Think about the shape of this curve: Cost is huge for large x (a tray!) and for small x (a tall tube!) In between, there should be a minimum.