Hmm. The most I can do with it is say 216 = Pi*r^2*h right? And then differentiate, but with respect to what. And then I get confused with the corner thing.
Suppose that a manufacterer is wanting to produce cans that'll hold 216 cubic inches that are in the form of a right circular cylinder. Determine the dimensions (radius of an end and height) of the can that'll use the least amount of material. Assume that circular ends are cut out of squares, with corner portions wasted, and further assume that the sides are made from rectangles, and there is no waste.
You need:
V = 216 cubic inches ... since V = pi*r^2*h, if we solve this for h, we get h = V/(pi*r^2) ... plugging in V = 216,
h = 216/(pi*r^2)
The diameter of the circle equals the side of the square being cut: s = D = 2*r
The area of the square sheet is: s^2 = 4*r^2
The circumference of the circle equals the the length of the rectangle being cut: L = C = 2*pi*r
The area of the rectangle making the side of the can is L*h = 2*pi*r*(216/(pi*r^2)) = 432/r ... got this after plugging in h = 216/(pi*r^2)
The total area for the material used for the can is: (2 times the area of one circle plus the area of the side sheet)
A = 2(4*r^2) + 432/r = 8r^2+432/r
Differeniting this, we get
A' = 16r - 432/r^2 ... Set A' equal to 0
0 = 16r - 432/r^2
432/r^2 = 16r
r^3 = 432/16 = 27
r = 3
plug r back into h = 216/(pi*r^2)
h = 216/(pi*3^2)
h = 216/(9pi)
h = 24/pi