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

a manufacturer of printed circuit boards uses exponential smoothing with trend to forecast monthly demand of its product. at the end of december, the company wishes to forecast sales for january. the estimate of trend through november has been 200 additional boards sold per month. average sales have been around 1000 units per month. the demand for december was 1100 units. the company uses α = 0.20 and β = 0.10. make a forecast including trend for the month of january.