What is the extreme f(x,y)=xe^{-30x^2-15y^2} ?

The extreme f(x,y)=xe^{-30x^2-15y^2} is

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extreme f(x,y)=xe^{-30x^2-15y^2}

extreme points

f (x, y) = x e^{- 30 x^{2} - 15 y^{2}}

Minimum (- \frac{1}{2 15}, 0), Maximum (\frac{1}{2 15}, 0), Minimum (- \frac{1}{2 15}, 0), Maximum (\frac{1}{2 15}, 0)

Solution steps

Find the critical points:

(- \frac{1}{2 15}, 0), (\frac{1}{2 15}, 0), (- \frac{1}{2 15}, 0), (\frac{1}{2 15}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 30 x (- 30 e^{- 30 x^{2} - 15 y^{2}} y^{2} + e^{- 30 x^{2} - 15 y^{2}})

D (x, y) = - 30 x (3600 e^{- 15 (y^{2} + 2 x^{2})} x^{3} - 180 e^{- 15 (y^{2} + 2 x^{2})} x) (- 30 e^{- 30 x^{2} - 15 y^{2}} y^{2} + e^{- 30 x^{2} - 15 y^{2}}) - 900 e^{- 30 (y^{2} + 2 x^{2})} y^{2} (215 x + 1)^{2} (215 x - 1)^{2}

Check the sign of at each critical point

Check critical point

(- \frac{1}{2 15}, 0) :

Minimum

Check critical point

(\frac{1}{2 15}, 0) :

Maximum

Check critical point

(- \frac{1}{2 15}, 0) :

Minimum

Check critical point

(\frac{1}{2 15}, 0) :

Maximum

Minimum (- \frac{1}{2 15}, 0), Maximum (\frac{1}{2 15}, 0), Minimum (- \frac{1}{2 15}, 0), Maximum (\frac{1}{2 15}, 0)

What is the extreme f(x,y)=xe^{-30x^2-15y^2} ?
The extreme f(x,y)=xe^{-30x^2-15y^2} is