Optimization Question

Question

A cylinder-shaped can need to be constructed to hold 600 cubic centimeters of soup. The material for the sides of the can costs 0.03 cents per square centimeter. The material for the top and bottom of the can needs 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:?Incorrect  

Height of the can:?  

Minimum cost:? cents

Answer 1

v=600 cm^3

= hpir^2

h=600/pir^2

c=.06(2pir^2)+ .03(2pirh)= .06(2pir^2+pi(600r/pir^2)

=.12pir^2 +36/r

c'(r)=.24pir -36/r^2=0

.24pir^3=36

r^3=36/,24pi

r=(36/.24pi)^(1/3)=(150/pi)^(1/3)= about 3.628 cm = radius

h=600/pir^2= 600/pi(150/pi)^(2/3)=about 14.51 cm= height

min cost = .12pir^2+36/r= about .12pi(3.628^2) +36/3.628

= about 14.88 cents

no guarantees answers are error free, but it's the basic optimization method

get cost in terms of one variable, take the derivative, set =0

But

the height = twice the diameter= 4 times the radius. h=4r

the side costs twice the top and bottom

leading to a much easier, non-calculus, calculation of the radius

600 = hpir^2 =4rpir^2 = 4pir^3

r^3 = 600/4pi

r = (600/4pi)^(1/3) =about 3.628 cm

h=4r = 4(3.628)= 14.512 cm

if you get the same answer with two different methods

that suggests virtual certainty the answer is correct