A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must have a volume of 1600 cubic feet. The hemispherical ends cost twice as much per square foot of surface area as the sides. Find the dimensions that will minimize the cost. (Round your answers to three decimal places.)

Answer 1

what, no effort since you last [posted this problem? Most of it is just geometry.

If the radius is r and the cylinder length is h, then

4/3 π r^3 + πr^2 h = 1600
so
h = (4800 - 4πr^3)/(3πr^2)
= 1600/πr^2 - 4r/3

If the sides cost $1/ft^2, then the ends cost $2/ft^2, so the cost is

c = 2*4πr^2 + 2πrh
= 2*4πr^2 + 2πr(1600/πr^2 - 4r/3)
= 3200/r + 16π/3 r^2

so, to minimize cost, find where dc/dr = 0

dc/dr = -3200/r^2 + 32πr/3
= (32πr^3-9600)/3r^2
dc/dr=0 when
32πr^3 = 9600
r^3 = 300/π

r = 4.57
h = 18.28

Answer 2

total volume = cylinder + 2(hemispheres)
= πr^2 h + (4/3)π r^3
1600 = πr^2 h + (4/3)π r^3
4800 = 3πr^2 h + 4π r^3
h = (4800 - 4π r^3)/(3π r^2)

cost = 2(4π r^2) + 2π rh
= 8π r^2 + 2πr(4800 - 4π r^3)/(3π r^2)
= (16/3)π r^2 + 3200/r , not showing the simplification

d(cost)/dr = (32/3)πr - 3200/r^2
= 0 for a max/min of cost

32πr/3 = 3200/r^2
32π r^3 = 9600
r^3 = 9600/(32π) = 300/π
r = appr 4.57 ft
then
h = 18.283

state conclusion, but first check my arithmetic

Answer 3

*whew* what a relief that we agreed!!