What is the extreme f(x,y)=x^2y^2+2xy ?

The extreme f(x,y)=x^2y^2+2xy is Saddle(0,0)

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extreme f(x,y)=x^2y^2+2xy

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 x^{2}

D (x, y) = - 12 x^{2} y^{2} - 16 x y - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 4 x^{2} - 5 x - 2

f (x, y) = x y - 2 x - 4 y

f (x) = 4 x^{2} - 5 x + 1

f (x) = x^{3} - 4 x^{2} - 3 x + 6

What is the extreme f(x,y)=x^2y^2+2xy ?
The extreme f(x,y)=x^2y^2+2xy is Saddle(0,0)