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.