An aluminium manufacturing company is working on designs for soup cans. In one designthey are going to use the same gauge metal for the whole can, it costs $0.002 per squarecentimeter of aluminum. If the can must hold 430ml of soup what dimensions will minimize the cost of the can. (1ml?

the final dimensions are R = 0.409 m and L = 0.818 m

Explanation:

Since the cost is proportional to the area, then we should find the can with volume V and minimum area. the volume is

V = π*L*R²

where L is length, R is radius

then since the area of the can = area of the sides + area of the cap and bottom.

A = 2*π*R*L + 2*π*R², V = π*L*R² → L = V / (π*R²)

A = 2*π*R * V / (π*R²) + 2*π*R² = 2*V/R + 2*π*R²

then we find the minimum area when the derivative of the area with respect to R is 0:

dA/dR = - 2*V/R² + 4*π*R = 0 → R = ∛[V / (2π) ]

replacing values

R = ∛[V / (2π) ] = ∛[0.430 m³ / (2π) ] = 0.409 m

thus

L = V / (π*R²) = 0.430 m³ / (π * (0.409 m) ²) = 0.818 m

therefore the final dimensions are

R = 0.409 m and L = 0.818 m