Applied Maximum and Minimum Problems

Question

A cylinder-shaped can must be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs 0.03 cents per square centimeter. The fabric for the top and bottom of the can need to be thicker and costs 0.06 cents per square centimeter. Find the dimensions for the can that will minimize production costs.

To minimize the cost of the can:

The radius of the can: ____

Height of the can: ____

Minimum Cost: ____ cents

Answer 1

400 = hpir^2

h = 400/pir^2

Surface Area

= area of top&bottom circles + pi(diameter)(height)

= pir^2 + pir^2 + pidh with d=2r

= 2pir^2 + 2pirh

= 2pir^2 +2pir(400/pir^2)

Cost = .06(2)pi(r^2) + .03(2)(400/r)

(taking the problem's costs as $.06 and $.03 = 6 cents and 3 cents)

C(r) = .12pir^2 +24/r

take the derivative and set equal to zero

C'(r) = .24pir -24/r^2 = 0

multiply by r^2

.24pir^3 -24 =0

.24pir^3 =24

.01pir^3 = 1

pir^3 = 100

r^3 = 100/pi

r = cube root of 100/pi

r = about cube root of 100/3.1416

r = 31.831^(1/3)

r =radius = about 3.169 cm

h=height = 400/pir^2 = about 12.677 cm

Minimum Cost =

= C(r)= .12pir^2+24/r

= C(3.169)

= .12(3.1416)(3.169^2) + 24/3.169

= 3.786 + 7.573

= $11.36

= 1,136 cents per can