MINIMIZATION Math Problem! How do I solve this... see details!?

Question

Assume that the box has a square base, must contain 32,000 cubic centimeters and that the cost of the material for the bottom is THREE times the cost of the material for the sides. Find the dimensions of the box that has the least materials cost.

Answer 1

The volume of a box with a square base is x²h (where x is the dimension of the square and h is the height of the box).

So 32000 = x²h --> h = 32000/x²

You are minimizing the cost function which is going to be as follows:

The 4 sides each have area xh and there are 4, so the total surface area is 4xh = 4x*(32000/x²) = 128000/x.

The bottom of the box has area x² (I am assuming this is an open box with no top because you don't mention the cost of the top, only the bottom).

So the total cost C(x) = 128000/x + 3x²

to minimize, you take the derivative and find critical value(s):

C'(x) = -128000/x² + 6x
0 = -128000/x² + 6x
0 = -128000 + 6x^3
6x^3 = 128000
x^3 = 21333.33
x = 27.73 cm and h = 128000/x² = 166.41 cm

{note: 0 is also a critical value because it causes C'(x) to be undefined, but the answer is obviously not x = 0. 27.73 is in fact a minimum when you look at the sign of C'(x) on either side of this critical value}

Answer 2

V= x^2*y=32,000cm^3
Cost = 3x^2*Price +4xy *Price
Price/cm^2
y= 32,000/x^2
Cost = Price( 3x^2 +4*32,000/x)
Derivate
6x-128,000/x^2= 0 so x=(128,000 /6)^1/3 = (64,000/3)^1/3 =27.7345 cm
y=41.6017 cm