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

1 answer:

3 0

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.

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