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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 :

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 $

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