Find two numbers x and y such that their sum is 60 and x2y is maximized.

X + y = 60
P = x^2 y

From first  equation y = 60 - x

P = x^2(60 - x)
P = 60x^2 - x^3
Finding the derivative of P and equating to zero:-
P' = 120x - 3x^2 = 0
 -3x ( x - 40) = 0
 x = 40 
This makes y = 20
So maximum values of x^2y  = 40^2 * 20 =  32,000.