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.

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jtellezd jtellezd · Answer 1 · 2019-12-07T01:13:37+01:00

Answer:

x = 2,94 ft

h = 0,88 ft

C(min) = 12,23 $

Step-by-step explanation:

Let x be radius of the base and h th hight of th cylinder then

Area of the bottom A₁ = π*x²

And cost of the bottom C₁ = 0,15*π*x²

Lateral area A₂

A₂ = 2*π*x*h but V = 24 ft³ V = π*x²*h h = V/ π*x²

A₂ = 2*π*x* (24/ π*x²)

A₂ = 48 /x

Cost of area A₂

C₂ = 0,05 * 48/x

C₂ = 2,4/x

Total cost C C = C₁ + C₂

C(x) = 0,15*π*x² + 24/x

Taking derivatives on both sides of the equation

C´(x) = 0,3**π*x - 24/x²

C´(x) = 0 0,3**π*x - 24/x² = 0

0,942 x³ = 24

x³ = 25,5

x = 2,94 ft and h = V/π*x² h = 24 /3,14* (2,94)²

h = 0,88 ft

C(min) = 0,15 * 3,14 * (2,94)² + 24/2,94

C(min) = 4,07 + 8,16

C(min) = 12,23 $