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

The extreme f(x,y)=3x^3-12xy+y^3 is Saddle(0,0),Minimum(4/(\sqrt[3]{9)},(4*9^{1/3})/3)

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

extreme

Saddle (0, 0), Minimum (\frac{4}{3 9}, \frac{4 \cdot 9 ^{\frac{1}{3}}}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{4}{3 9}, \frac{4 \cdot 9 ^{\frac{1}{3}}}{3})

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

D (x, y) = 108 x y - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{4}{3 9}, \frac{4 \cdot 9 ^{\frac{1}{3}}}{3}) :

Minimum

Saddle (0, 0), Minimum (\frac{4}{3 9}, \frac{4 \cdot 9 ^{\frac{1}{3}}}{3})

What is the extreme f(x,y)=3x^3-12xy+y^3 ?
The extreme f(x,y)=3x^3-12xy+y^3 is Saddle(0,0),Minimum(4/(\sqrt[3]{9)},(4*9^{1/3})/3)