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

The extreme f(x,y)=yx^2-3xye^{-xy} is Saddle(0,0),Saddle(3,0)

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extreme f(x,y)=yx^2-3xye^{-xy}

extreme points

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

Saddle (0, 0), Saddle (3, 0)

Solution steps

Find the critical points:

(0, 0), (3, 0)

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

D (x, y) = 18 e^{- 2 x y} x^{3} y^{3} + 6 e^{- x y} x^{3} y^{2} - 63 e^{- 2 x y} x^{2} y^{2} - 24 e^{- x y} x^{2} y - 4 x^{2} + 54 e^{- 2 x y} x y + 12 e^{- x y} x - 9 e^{- 2 x y}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(3, 0) :

Saddle

Saddle (0, 0), Saddle (3, 0)

f (x, y) = x^{3} - 9 x y - 4 y^{2}

y = 4 - 6 x^{2}

\frac{x ^{2} + 2}{x - 3}

f (x) = 5 + \frac{48}{x} + 3 x^{2}

E (x, y) = 15 x y - 22 x y

What is the extreme f(x,y)=yx^2-3xye^{-xy} ?
The extreme f(x,y)=yx^2-3xye^{-xy} is Saddle(0,0),Saddle(3,0)