MinMax06 Minumum cost to construct a cylinder.

Find the minimum cost to construct a cylindrical container if material for the top and bottom costs 4 cents per square inch and material for the sides costs 3 cents per square inch. The container is to have volume 100 cubic inches.

Draw a cylinder of radius r and height h. The area of the top and bottom is pr². The area of the side is (2pr)h . Imagine the side as a piece 2pr long, the circumference of the container, and h high.

The defining equation is the cost equation which in words is 4 cents times the area of the top and bottom plus 3 cents times the area of the side.

C = 4(pr² + pr²) + 3(2prh) = 8pr² + 6prh

The constraint is that the volume must be 100 cubic inches. The volume of a cylindrical container is the area of the bottom, pr², times the height, h: V = pr²h .

Set V = 100, solve for h, and substitute into the defining equation: 100 = pr²h or h = 100/(pr²) and

C = 8pr² + 6pr[100/(pr²)] = 8pr² + 600/r = 8pr² + 600r^-1

The first derivative of C is: C' = 16pr - 600r^-2 and setting C' = 0 produces

16pr - 600/r² = 0 or r³ = 600/(16p) and r =(600/16p)^1/3 ť 2.3

The second derivative of C is: C'' = 16p + 1200/r³

C'' is positive for all positive r indicating a minimum for the curve.

Substituting the r for zero slope back into the constraint equation 100 = pr²h produces

100 = p[600/(16p)]2/3h or h = (100/p) (16p/600)^2/3 ť 6.1