A manufacturer of aluminum cans has budgeted 20ð square centimeters of material for each can. What is the maximum volume that such a can could hold?

Answer 1

Radius = r
Height = h
Volume, V = πr²h
Surface area, S = 2πr²+ 2πrh
=2πr(r+h)

Given S=20π, therefore
2πr(r+h) = 20π, or
r(r+h)=10
h=(10/r-r)

V(r) = πr²h
=πr²(10/r-r)
=π(10r-r³)

Since V(r) is now expressed in terms of r, calculate the derivative V'(r).
Equating V'(r)=0 gives the value of r that maximizes or minimizes the volume.
I get r=1.8 approx.

Calculate the second derivative V"(r) of the volume with respect to r, and evaluate V"(1.8) to make sure it is negative, which implies V(1.8) is a maximum.

I get Vmax=38 approx.

Check my work and my thinking.

Answer 2

let the radius of the can be r cm and its height be h cm
2pi(r^2) + 2pi(rh) = 20ð

(I don't know what the symbol at the end of 20ð is, I will continue assuming it is 200. If it is anything else, you will have to adjust the calculations)
2pi(r^2) + 2pi(rh) = 200
h = 100/(pir) - r

volume = pi(r^2)h
= pir^2(100/(pir) - r)
= 100r - pr^3
d(volume)/dr = 100 - 2pir^2 = 0 for a max/min of volume
2pir^2 = 100
r^2 = 50/pi
r = 3.989
then h = 3.989

which confirms the concept that the largest volume of a can using the least material is obtained when the height and radius are the same.