Find two non-negative numbers x and y whose sum is 100 and for which x^2y is a maximum

 x + y = 100 ==> x = (100 - y) 
P = xy^2 = (100 - y)y^2 = 100y^2 - y^3 
P ' = 200 y - 3y^2 
P" = 200 - 6y 

P' = 0 when 200 y - 3y^2 = 0 
y(200 - 3y) = 0 
so either y = 0 or 200 - 3y = 0 
y = 0 or y = 66.667 
and since x = 100 - y, then if y = 0, then x = 100 
or if y = 66.667, then x = 33.333 

by inspection, if y = 0, the product is 0 (a minimum) 

so (33.333 , 66.667) is the maximum 
P " (66.667) < 0, confirming this is a max 

The solution to the problem is as follows:

x+y = 100

f(x) = x²y = x²(100-x) = 100x² - x³ 

f'(x) = 200x - x² = 0 => x = 0, 200 

x = 200 

y = -100 

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