What is the extreme f(x)=x^3+y^3-12xy ?

The extreme f(x)=x^3+y^3-12xy is Saddle(0,0),Minimum(4,4)

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extreme f(x)=x^3+y^3-12xy

extreme

Saddle (0, 0), Minimum (4, 4)

Solution steps

Find the critical points:

(0, 0), (4, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, 4) :

Minimum

Saddle (0, 0), Minimum (4, 4)

What is the extreme f(x)=x^3+y^3-12xy ?
The extreme f(x)=x^3+y^3-12xy is Saddle(0,0),Minimum(4,4)