Maximization/Minimization Problem help

Hey all, so I have a Basic Skills Test for multivariable calculus due at midnight, and I'm stuck on this one optimization problem. Here it is:

A rectangular storage container with an open top is to have a volume of 10 cubic meters. The length of this base is twice the width. Material for the base costs $5 per square meter. Material for the sides costs $3 per square meter. Find the cost of materials for the cheapest such container.

What I've figured out is that total price would be P=10w^2 + 18hw, the volume is V=2hw^2=10. But without any kind of reference to height, I'm still stuck with a variable.

Thanks in advance!

You first begin with the base of the box costing 2x^2 if x is the width. The total volume of the box would be equal to 2x^2 * Y where Y is the height. Because you know the volume is equal to 10, you can set 2x^2*Y=10. Then you can have Y in terms of x. You simply divide both sides by 2x^2, so you get Y=10/(2x^2). So, you get the area of all of the parts of the box which are 2x^2+2(5/x)+2(10/x) which is the total area of the box. Then apply the prices, so we get 10x^2+90/x=P. We take the derivative, and we get 20x-(90/x^2)='P. We then set it to zero and solve for x. When you solve for x, it comes out to be 4.5^(1/3)=x. Plug in this to the original equation of 10x^2+90/x and we get around $81.77.

Solve your volume equation for h, then plug it into the cost equation.

2hw^2 = 10
h = 5 / w^2

P = 10w^2 + 18w(5/w^2) = 10w^2 + 90/w

Take the derivative and set it equal to zero.

P' = 20w - 90/w^2 = 0
20w = 90/w^2
w^3 = 4.5
w = 1.65096

P = $81.77

w^2 = 5/h
h=1.834
w=1.651
l=3.302