Optimization Problem...?

Question

A closed cylindrical can is to hold 1 liter of liquid. How should we choose the height and radius to minimize the amount of material needed to manufacture the can?
Thanks a bunch!

Answer 1

material needed=2pir(h+r)
volume=pir^2h
pir^2h=1000cc (given)
h=1000/(pir^2)
substituing h in the equation 2pir(h+r)
material=2pir(1000/pir^2+r)
=2pir*1000/pir^2+2pir^2
dA/dt=-2000/r^2+4pir
setting this to zero
-2000+4pir^3=0
4pir^3=2000
r^3=2000/4pi
r=cube root of 500/3.14
=5.4 approx
substituting
h=1000/(3.14*5.4*5.4)
=10.9 aapprox
so the radius is 5.4 cm approx
and the height 10.9 approx

Answer 2

The total area of material for the can is twice the area of the base (πr^2) plus the area of the sides (2πrh). The total surface area is then A = 2πr^2+πrh. The volume of the can is V = πr^2h. Solve this for h: h = V/πr^2 and plug into the eq for A:

A = 2πr^2 + πrV/πr^2 = 2πr^2 + V/r

differentiate wrt r and set to 0

A' = 4πr - V/r^2 = 0
4πr^3 = V
r = [V/4π]^1/3

Since h = V/πr^2, h = V/π[V/4π]^2/3 = V^1/3/π(1/4π)^2/3
h = 4^2/3 * (V/π)^1/3 = (16V/π)^1/3

Answer 3

V = πr²h = 1000 cm³
h = 1000/πr²

S = 2πr² + 2πrh
= 2πr² + 2πr * 1000/πr²
= 2πr² + 2000/r

dS/dr = 4πr - 2000/r²
= 0 for stationary points

ie 4πr³ = 2000
so r³ = 500/π
So r = (500/π)^⅓
≈ 5.42 cm
h = 1000/πr²
≈ 10.84 cm

Check V = πr²h
≈ π * 5.42² *10.84 = 1000.4 cm³ (error is truncation error)