Calculus help? Applied Maximum and Minimum problems.

Problem: A closed rectangular container with a square base is to have a volume of 2,000 cm^3. It costs twice as much per square centimeter for the top and bottom as it does for the sides. Find the dimensions of the container of least cost.

I'll need to sketch the objects in these problems so I'm assuming that this is merely a rectangular box? Like this one:
http://www.onlineconversion.com/images/o…

Let L = length, W = width, H = height. Since the base is a square, W = L.

volume: HL^2 = 2000

The sides have area HL and HW (=HL) and the top/bottom have area LW (=L^2). So total cost Y is

Y := 2*HL + 2*HL + 2*2*L^2 = 4LH + 4L^2

Substitute H = 2000/L^2 into the second equation, you get

Y = 4L(2000/L^2) + 4L^2 = 8000/L + 4L^2

To find the maximum, take the derivative and set to zero:

dY/dL = -8000/L^2 + 8L = 0
==> 8L = 8000/L^2
==> L^3 = 1000
==> L = 10

Hence L = 10, W= 10, and H = 2000/L^2 = 20.