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

The extreme f(x,y)=x^3-xy+3y^2 is Minimum(1/18 , 1/108),Saddle(0,0)

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

extreme points

f (x, y) = x^{3} - x y + 3 y^{2}

Minimum (\frac{1}{18}, \frac{1}{108}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{18}, \frac{1}{108}), (0, 0)

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

D (x, y) = 36 x - 1

Check the sign of at each critical point

Check critical point

(\frac{1}{18}, \frac{1}{108}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{18}, \frac{1}{108}), Saddle (0, 0)

f (x) = 4 x^{4} + y^{2} - 2 x y + 1

f (x) = x^{3} - 3 x^{2} - 24 x + 5

f (x, y) = 12 x y - x^{3} - 6 y^{2}

x^{3} - x^{2}

f (x, y) = x y + \frac{27}{x} + \frac{27}{y}

What is the extreme f(x,y)=x^3-xy+3y^2 ?
The extreme f(x,y)=x^3-xy+3y^2 is Minimum(1/18 , 1/108),Saddle(0,0)