A 2746 ft^3 tank (with a top) is to be built by welding four sides of height h to a square base of width x from sheet metal, and then a top the same size as the base is put on.

Question

The tank must also sit on a cement base 1 ft deep which should be the same size as the base of the tank. If the metal costs 24 dollars per square foot and the cement costs 13 dollars per cubic foot, what are the dimensions of the tank which minimizes the cost of the project?

width=

length=

height=

Answer 1

Let:

x = length and width of the square base of the tank

h = height of the tank

V_tank = Volume of the tank = 2746 ft³

SA_tank = surface area of the tank

V_cement = Volume of the cement base

V_tank = x²h = 2746

h = 2746/x²

Since the tank is rectangular prism, then the surface area of the tank is:

SA_tank = 2x² + 4xh

substitution:

SA_tank = 2x² + 4x(2746/x²)

SA_tank = 2x² + 10984/x

Volume of the cement base is:

V_cement = x²(1) = x²

The cost of the metal is 24*SA_tank and the cost of the cement is 13*V_cement. Therefore the total cost (C) is:

C = 24(2x² + 10984/x) + 13x²

Simplify:

C = 61x² + 263616/x

Get the derivative of C:

dC/dx = 122x - 263616/x²

Make dC/dx = 0 to get the local minimum.

0 = 122x - 263616/x²

0 = (122x³ - 263616)/x²

0 = 122x³ - 263616

x ≈ 12.928 ft. (both length and width because it's a square.)



For the height:

h = 2746/(12.928...)²

h ≈ 16.430 ft.