A cylindrical metal container, open at the top, is to have a capacity of 24 cu. in. The cost of material used for the bottom of the container is $0.15/sq. in., and the cost of the material used for the curved part is $0.05/sq. in. Find the dimensions that will minimize the cost of the material, and find the minimum cost.
Accepted Solution
A:
Answer: Dimensions:X = radius of base = 2.76 in L = height = 1 inMinimum cost = 4.456 $Step-by-step explanation:We have:V 24 in³ V = πx²L where x radius of base of the cylinder and h the height of the cylinder.then L = V/πx² L = 24/πx²Area (x) = Base area + lateral areaA(x) = πx² + 2πxL ⇒ A(x) = πx² + 2πx(24/πx²) ⇒A(x) = πx² +48/xTaking derivatives:A´(x) = 2πx - 48/x² A´(x) = 0 6.28x² - 48 = 0 x² = 48 / 6,28 x² = √7.64 x = 2.76 radius of base (in)and the height L = 24/(3,14)*(2.76)² L = 1 incost : C($) =(0,15) * (2.76)²*3.14 + 2*(3.14)(2.76) *1 *0.05C($) = 3.59 + 0.866 C($) = 4.456 $