What is the extreme f(x,y)=x^3-2y^2-6xy+4 ?

The extreme f(x,y)=x^3-2y^2-6xy+4 is Maximum(-3, 9/2),Saddle(0,0)

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extreme f(x,y)=x^3-2y^2-6xy+4

extreme

Maximum (- 3, \frac{9}{2}), Saddle (0, 0)

Solution steps

Find the critical points:

(- 3, \frac{9}{2}), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = - 24 x - 36

Check the sign of at each critical point

Check critical point

(- 3, \frac{9}{2}) :

Maximum

Check critical point

(0, 0) :

Saddle

Maximum (- 3, \frac{9}{2}), Saddle (0, 0)

What is the extreme f(x,y)=x^3-2y^2-6xy+4 ?
The extreme f(x,y)=x^3-2y^2-6xy+4 is Maximum(-3, 9/2),Saddle(0,0)