V = pi r^2 h = 500
so h = 500/(pi r^2)
area of side = 2 pi r h
area of top and bottom togethr = 2 (pi r^2)
=2 pi r^2
cost = C = .11(2 pi r^2) +.06(2 pi r h)
C = .22 pi r^2 + .12 pi r [500/(pi r^2)]
C = .22 pi r^2 + 60/r
C = .691 r^2 + 60/r
I need to use calculus to find the maximum or minimum
dC/dr = 0 at extreme = 1.38 r -60/r^2
60/r^2 = 1.38 r
r^3 = 43.4
r = 3.51 cm for minimum
Could you help me with this math question:
A can in the shape of a right circular cylinder is required to have a volume of 500 cubic centimeters. The top and bottom are made of material that costs 11 cents per square centimeter, while the sides are made of a material that costs 6 cents per square centimeter. Express the total cost C of the material as a function of the radius r of the cylinder. Find the right side of the equation. Express the cost in dollars. For what value of r is the cost C a minimum?
2 answers
C(r) = .11(2pi r^2) + .06(2pi rh)
since pi r^2 h = 500,
h = 500/(pi r^2)
C(r) = .11(2pi r^2) + .06(2pi r(500/pi r^2))
C(r) = .22 pi r^2 + 60/r
For C to be a minimum,
C'(r) = .44pi r - 60/r^2 = 0
.44pi r^3 - 60 = 0
r^3 = 60/.44pi
r = 3.51
since pi r^2 h = 500,
h = 500/(pi r^2)
C(r) = .11(2pi r^2) + .06(2pi r(500/pi r^2))
C(r) = .22 pi r^2 + 60/r
For C to be a minimum,
C'(r) = .44pi r - 60/r^2 = 0
.44pi r^3 - 60 = 0
r^3 = 60/.44pi
r = 3.51