What is the extreme f(x,y)=-2xye^{-x^2-y^2} ?

The extreme f(x,y)=-2xye^{-x^2-y^2} is

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extreme f(x,y)=-2xye^{-x^2-y^2}

extreme points

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

Saddle (0, 0), Minimum (- \frac{1}{2}, - \frac{1}{2}), Maximum (\frac{1}{2}, - \frac{1}{2}), Maximum (- \frac{1}{2}, \frac{1}{2}), Minimum (\frac{1}{2}, \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{2}, - \frac{1}{2}), (\frac{1}{2}, - \frac{1}{2}), (- \frac{1}{2}, \frac{1}{2}), (\frac{1}{2}, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2 x (4 e^{- x^{2} - y^{2}} y^{3} - 6 e^{- x^{2} - y^{2}} y)

D (x, y) = - 32 e^{- 2 x^{2} - 2 y^{2}} x^{4} y^{2} - 16 e^{- 2 x^{2} - 2 y^{2}} x^{4} - 32 e^{- 2 x^{2} - 2 y^{2}} x^{2} y^{4} + 80 e^{- 2 x^{2} - 2 y^{2}} x^{2} y^{2} + 16 e^{- 2 x^{2} - 2 y^{2}} x^{2} - 16 e^{- 2 x^{2} - 2 y^{2}} y^{4} + 16 e^{- 2 x^{2} - 2 y^{2}} y^{2} - 4 e^{- 2 x^{2} - 2 y^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{2}, - \frac{1}{2}) :

Minimum

Check critical point

(\frac{1}{2}, - \frac{1}{2}) :

Maximum

Check critical point

(- \frac{1}{2}, \frac{1}{2}) :

Maximum

Check critical point

(\frac{1}{2}, \frac{1}{2}) :

Minimum

Saddle (0, 0), Minimum (- \frac{1}{2}, - \frac{1}{2}), Maximum (\frac{1}{2}, - \frac{1}{2}), Maximum (- \frac{1}{2}, \frac{1}{2}), Minimum (\frac{1}{2}, \frac{1}{2})

f (x, y) = x y + x^{2} y - x y^{2}

f (x) = x^{3} + 6 x^{2} + 12 x + 2

f (x) = - 0.002 x^{2} + 3.2 x - 60

f (x) = x^{3} + 6 x^{2} + 12 x + 3

f (x) = 7 x^{4} - 28 x^{3} + 5

What is the extreme f(x,y)=-2xye^{-x^2-y^2} ?
The extreme f(x,y)=-2xye^{-x^2-y^2} is