Question

An aluminum can is to be constructed to contain 2300 cm^3 of liquid. Let r and h be the radius of the base and the height of the can respectively.

Answer 1

A) Volume = V = 2200 = πR²h

h = 2200/(πR²)

B) Area = A = 2πRh + 2πR² = 2πR(h+R)

A = 2πR(R + 2200/πR²)

A = 2πR² + 4400/R

C) dA/dR = 4πR - 4400/R² = 0

πR³ - 1100 = 0

R³ = 1100/π

R ≈ 7.048 cm

h = 2200/(πR²) ≈ 14.096 cm

The dimensions which use the minimum amount of material for the container are:

R ≈ 7.048 cm

h ≈ 14.096 cm

(Note: the height is exactly equal to the diameter of the container or twice the radius)