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.

Answer: Dimensions:

X = radius of base = 2.76 in L = height = 1 in

Minimum 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/πx²

Area (x) = Base area + lateral area

A (x) = πx² + 2πxL ⇒ A (x) = πx² + 2πx (24/πx²) ⇒A (x) = πx² + 48/x

Taking 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 in

cost : C ($) = (0,15) * (2.76) ²*3.14 + 2 * (3.14) (2.76) * 1 * 0.05

C ($) = 3.59 + 0.866

C ($) = 4.456 $