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

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