The volume of a large soup can is 1L. The cost of the top is $0.002/cm^2 while the sides and bottom cost $0.001/cm^2.Find the minimum cost to pruduce the can given that the radius must be between 5cm and 10cm for shipping purposes.

Answer 1

let the radius of the can be r cm and its height be h cm

given: πr^2 h = 1000
h = 1000/(πr^2)

cost = .002(top+bottom) + .001(sleeve)
= .002(2πr^2) + .001(2πrh)
= .004πr^2 + .002πr(1000/πr^2 )
= .004πr^2 + .002(1000/r)
= .002 [2πr^2 + 1000/r ]

d(cost)/dr
= .002 (4πr - 1000/r^20
= 0 for a min of cost

4πr = 1000/r^2
4πr^3 = 1000
r^3 = 250/π
r = 4.30127..
then h = 17.205

but the radius must be at least 5 cm, so that's it
r = 5, h = 12.73

minimum cost = .02[2π(25) + 1000/5]
= 7.14 cents to make a can

check:
at "real" best answer:
cost = .02(2π(4.30127)^2 + 1000/4.30127)
= 5.812

take case of slightly off that,
r = 4.3, h = 17.2
cost = 6.97 , higher

r = 4.4, h = 16.44
cost = 6.97, higher

at restricted value of
r = 5, h = 1.73
cost = 7.14