Min max word problem? How would I solve the following problem.
Problem:
A rectangular warehouse will hold 8000 square feet of floor space and will be separated into two rectangular rooms by an interior wall. The cost of the exterior wall is 120 per linear foot and the cost of the interior wall is 100 per linear foot. Find the dimensions that will minimize the cost of building the warehouse. 
This is what I did
$xy=8000$ $y=\frac{8000}{x}$
exterior wall $2(x+y)(120)$
interior wall $(x)(100)$
$C=2(x+y)(120)+100x$
$240(x+y)+100x$
$340x+240(8000)x^{-1}$
$340x+\frac{1920000}{x}$
$c'(x)=340-\frac{1920000}{x^2}$
$\frac{340x^2-1920000}{x^2}$
$340x^2-1920000=0$
$x=75.15$
The minimum cost is approximatly $x=75.15$
 A: You're work looks good, but you need to solve for $y$, too.
And then you need to use your solutions for $x, y$ to compute the total cost using you $C$-equation, to calculate minimum cost.
So you want to find the dimensions of the rectangle: solutions for $x, y$
And with those values, compute the minimum cost: substitute $x$, $y$ into you cost equation to evaluate minimum cost: $c$

Summary from comments (deleted to clean up):
Try to keep your solution for $x$ as precise as possible, so you can have a precise figure to compute $y$: $x \approx 75.147,$ but $x^* = \frac{80}{17}\sqrt{255}$.
To solve for $y$: use the equation $$y = 8000/x = \dfrac{8000}{\frac{80\sqrt{255}}{17}} = \dfrac{100\cdot 17}{\sqrt{255}}$$
And finally, you need to show that (even though we know that the minimum must occur when $c'(x) = 0$, and need $x>0$, hence only one possible minimum exists, you should still justify that your solution: 
$x^*$ indeed gives a minimum: that $c'(x) < 0$ when $0 \lt x \lt x^*$ and $c'(x) > 0$ when $x>x^*$
