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
Best Answer
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.
Differentiate Cost with respect to x, set the derivative = 0 and solve...