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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.
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