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

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

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extreme f(x,y)=3xy-2x^3y^3

extreme

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 12 x^{3} y

D (x, y) = - 180 x^{4} y^{4} + 108 x^{2} y^{2} - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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