What is the extreme f(x,y)=-2x^3-2y^3-6xy+1 ?

The extreme f(x,y)=-2x^3-2y^3-6xy+1 is Saddle(0,0),Minimum(-1,-1)

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extreme f(x,y)=-2x^3-2y^3-6xy+1

extreme

Saddle (0, 0), Minimum (- 1, - 1)

Solution steps

Find the critical points:

(0, 0), (- 1, - 1)

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

D (x, y) = 144 x y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, - 1) :

Minimum

Saddle (0, 0), Minimum (- 1, - 1)

What is the extreme f(x,y)=-2x^3-2y^3-6xy+1 ?
The extreme f(x,y)=-2x^3-2y^3-6xy+1 is Saddle(0,0),Minimum(-1,-1)