Question

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.

Answer 1

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 $