if the sum of two numbers is 3 then the maximum value of the product of the first and the square of second is

hi,

let one no. be  a   (0=< a  = <3 )

  other no. will be 3- a

so, we have to maximize (3-a ) * a^2

 let p =  (3-a ) * a^2

  dp/da =  6a - 3a^2

            = 0

emplies,    either a= 0 , or

             a =2

d^2 p / da^2 =  6 - 6a

so, at a = 0 , we will have a min .  (d^2 p / da^2 >0 )

  d^3 p / da^3  =  -6   emplies that at a= 2 we will get a max.

 so the max. value will be 4.