What is the extreme f(x,y)=xe^{-3x^2-3y^2} ?

The extreme f(x,y)=xe^{-3x^2-3y^2} is

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extreme f(x,y)=xe^{-3x^2-3y^2}

extreme points

f (x, y) = x e^{- 3 x^{2} - 3 y^{2}}

Minimum (- \frac{1}{6}, 0), Maximum (\frac{1}{6}, 0), Minimum (- \frac{1}{6}, 0), Maximum (\frac{1}{6}, 0)

Solution steps

Find the critical points:

(- \frac{1}{6}, 0), (\frac{1}{6}, 0), (- \frac{1}{6}, 0), (\frac{1}{6}, 0)

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

D (x, y) = - 6 x (36 e^{- 3 (x^{2} + y^{2})} x^{3} - 18 e^{- 3 (x^{2} + y^{2})} x) (- 6 e^{- 3 x^{2} - 3 y^{2}} y^{2} + e^{- 3 x^{2} - 3 y^{2}}) - 36 e^{- 6 (x^{2} + y^{2})} y^{2} (6 x + 1)^{2} (6 x - 1)^{2}

Check the sign of at each critical point

Check critical point

(- \frac{1}{6}, 0) :

Minimum

Check critical point

(\frac{1}{6}, 0) :

Maximum

Check critical point

(- \frac{1}{6}, 0) :

Minimum

Check critical point

(\frac{1}{6}, 0) :

Maximum

Minimum (- \frac{1}{6}, 0), Maximum (\frac{1}{6}, 0), Minimum (- \frac{1}{6}, 0), Maximum (\frac{1}{6}, 0)

f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 2

\frac{1}{x - 2}

f (x) = - \frac{5 e ^{- x}}{5 - 3 x}

f (x, y) = 3 x^{2} - 2 x y + y^{2} - 8 y

f (x, y) = x^{2} + y^{2} - 6 y + 9

What is the extreme f(x,y)=xe^{-3x^2-3y^2} ?
The extreme f(x,y)=xe^{-3x^2-3y^2} is