MATH SOLVE

5 months ago

Q:
# A cylindrical metal container, open at the top. is to have a capacity of 24pi 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 ihe minimum cost.

Accepted Solution

A:

Answer:x = 2,94 ft h = 0,88 ftC(min) = 12,23 $Step-by-step explanation:Let x be radius of the base and h th hight of th cylinder thenArea of the bottom A₁ = π*x²And cost of the bottom C₁ = 0,15*π*x²Lateral area A₂A₂ = 2*π*x*h but V = 24 ft³ V = π*x²*h h = V/ π*x²A₂ = 2*π*x* (24/ π*x²)A₂ = 48 /xCost of area A₂C₂ = 0,05 * 48/x C₂ = 2,4/xTotal cost C C = C₁ + C₂ C(x) = 0,15*π*x² + 24/xTaking derivatives on both sides of the equationC´(x) = 0,3**π*x - 24/x²C´(x) = 0 0,3**π*x - 24/x² = 00,942 x³ = 24x³ = 25,5x = 2,94 ft and h = V/π*x² h = 24 /3,14* (2,94)²h = 0,88 ftC(min) = 0,15 * 3,14 * (2,94)² + 24/2,94C(min) = 4,07 + 8,16C(min) = 12,23 $