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: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 $