Posted by **JEZRELL** on Saturday, February 23, 2013 at 11:43pm.

A company wants to manufacture cylindrical aluminum cans with a value of 1000cm^3 (1 Liter) What radius and height of the can be to minimize the amount of aluminum used? Please help! This is very hard. Thanks.

- Elementary Analysis -
**Steve**, Sunday, February 24, 2013 at 6:47am
You know that

pi r^2 h = 1000, so

h = 1000/(pi r^2)

the surface area is

a = 2pi r^2 + 2pi r h

= 2pi r^2 + 2pi r * 1000/(pi r^2)

= 2pi r^2 + 2000/r

da/dr = 4pi r - 2000/r^2

= (4pi r^3 - 2000)/r^2

Since r > 0, da/dr=0 when

4pi r^3 = 2000

r^3 = 2000/4pi = 500/pi

r = 5 ∛(4/pi) ~= 5.42

go to wolframalpha.com and enter

plot y = 2pi r^2 + 2000/r where 3<r<8

to see where it is minimum

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