A cylinder shaped can needs to be constructed to hold 600 cubic centimeters of soup. The material for the sides of the can costs 0.03 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.05 cents per square centimeter. Find the dimensions for the can that will minimize production cos

the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm

Step-by-step explanation:

since the volume of a cylinder is

V= π*R²*L → L =V/ (π*R²)

the cost function is

Cost = cost of side material * side area  + cost of top and bottom material * top and bottom area

C = a* 2*π*R*L + b* 2*π*R²

replacing the value of L

C = a* 2*π*R* V/ (π*R²) + b* 2*π*R²  = a* 2*V/R + b* 2*π*R²

then the optimal radius for minimum cost can be found when the derivative of the cost with respect to the radius equals 0 , then

dC/dR = -2*a*V/R² + 4*π*b*R = 0

4*π*b*R = 2*a*V/R²

R³ = a*V/(2*π*b)

R=  ∛( a*V/(2*π*b))

replacing values

R=  ∛( a*V/(2*π*b)) = ∛(0.03$/cm² * 600 cm³ /(2*π* 0.05$/cm²) )= 3.85 cm

then

L =V/ (π*R²) = 600 cm³/(π*(3.85 cm)²) = 12.88 cm

therefore the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm

so then what are you asking?

step-by-step explanation: