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

The extreme f(xy)=x^3-y^2-3x-2y-2xy-3 is Saddle(1/3 ,-4/3),Maximum(-1,0)

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

extreme

Saddle (\frac{1}{3}, - \frac{4}{3}), Maximum (- 1, 0)

Solution steps

Find the critical points:

(\frac{1}{3}, - \frac{4}{3}), (- 1, 0)

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

D (x, y) = - 12 x - 4

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, - \frac{4}{3}) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Saddle (\frac{1}{3}, - \frac{4}{3}), Maximum (- 1, 0)

What is the extreme f(xy)=x^3-y^2-3x-2y-2xy-3 ?
The extreme f(xy)=x^3-y^2-3x-2y-2xy-3 is Saddle(1/3 ,-4/3),Maximum(-1,0)