Let r be the radius, h the height.
Cost=2K(2PIr^2)+ k*h*PI*(2r)
I will be happy to critique your work on this. A most excellent problem.
A cylindrical can is to be made to hold 100cm cubic.The material for its top and bottom cost twice as much per cm square as that for its side.
The objective of this problem is to find the radius and the height of this cylinder such that its cost is minimized.Denote the radius of this can by r and its height by h. Let k be the cost,expressed in cents,of 1 cm square of the side of this can.
1)Express the cost of the side and that two bases in terms of r, h and k.
2)Express the total cost of manufacturing this can in terms of r only.
3)Find r that minimizes the cost.
4)Deduce the corresponding h.
2 answers
we know volume = πr^2h = 100
h = 100/(πr^2)
k = 2(2πr^2) + 1(2πrh)
= 4πr^2 = 2πr(100/(πr^2))
= = 4πr^2 + 200/r
dk/dr = 8πr - 200/r^2 = 0 for a max/min of k
8πr = 200/r^2
r^3 = 200/(8π)
take it from here
h = 100/(πr^2)
k = 2(2πr^2) + 1(2πrh)
= 4πr^2 = 2πr(100/(πr^2))
= = 4πr^2 + 200/r
dk/dr = 8πr - 200/r^2 = 0 for a max/min of k
8πr = 200/r^2
r^3 = 200/(8π)
take it from here
4 answers