What is the extreme f(x,y)=xye^{(-x-y)} ?

The extreme f(x,y)=xye^{(-x-y)} is Maximum(1,1)

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extreme f(x,y)=xye^{(-x-y)}

extreme points

f (x, y) = x y e^{(- x - y)}

Maximum (1, 1)

Solution steps

Find the critical points:

(1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = x (e^{- x - y} y - 2 e^{- x - y})

D (x, y) = x y (e^{- x - y} x - 2 e^{- x - y}) (e^{- x - y} y - 2 e^{- x - y}) - e^{2 (- y - x)} (1 - x)^{2} (1 - y)^{2}

Check the sign of at each critical point

Check critical point

(1, 1) :

Maximum

Maximum (1, 1)

x^{2} - 6 x + 5

\frac{x}{x + 3}

\frac{1 - x ^{2}}{2 x + 1}

f (x) = \frac{4 x - 2}{x ^{2} - 10 x + 25}

f (x, y) = x y + 5 x - 5

What is the extreme f(x,y)=xye^{(-x-y)} ?
The extreme f(x,y)=xye^{(-x-y)} is Maximum(1,1)