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.
Answers
Answered by
5
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)
Similar questions