Calculus
posted by Rae .
A steel storage tank for propane gas is to be constructed in the shape of a right circular cylinder with a hempishere at each end. The cost per square foot of constructing the end pieces is twice that of contructing the cylindrical piece. If the desired capacity of the tank is 144 pi cubic feet, what dimensions will minimize the cost of construction?

The two hemispheres at the ends will make up one complete sphere
Let the radius of the sphere be r ft, and let the height of the cylindrical part be h
Volume = cylinder + sphere
πr^2 h + (4/3)πr^3 = 144π
3πr^2 h + 4πr^3 = 432π
3r^2 h + 4r^3 = 432
h = (432  4r^3)/(3r^2)
The cost of production is dependent on the surface area of material used
SA = 4π^2 + 2πrh
= 4πr^2 + 2π(432  4r^3)/(3r^2)
= 4πr^2 + 288π/r^2  (8/3)πr
d(SA)/dr = 8πr  576π/r^3  8π/3 = 0 for a min of SA
8r  576/r^3  8/3 = 0
times 3r^3
24r^4 1728  8r^3 = 0
3r^4  r^3  216 = 0
At this point I "cheated" a bit and ran it through Wolfram
http://www.wolframalpha.com/input/?i=3r%5E4++r%5E3++216+%3D+0
for a real solution of r=3, r = appr3 and 2 complex roots,
so the only usable answer is r = 3
the h = (432  108)/27 = 12
So the cylinder part should be 12 ft long and the radius of each of the semispheres should be 3 ft 
remember the sphere surface is twice as expensive as the cylinder which has no compound curvature so is easy to roll.
c = 2(4 pi r^2) + 1(2 pi r h)
c= 8 pi r^2 + 2 pi r h
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