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

The extreme f(x,y)=x^3-3xy+3y^2 is Minimum(1/2 , 1/4),Saddle(0,0)

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

extreme

Minimum (\frac{1}{2}, \frac{1}{4}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{2}, \frac{1}{4}), (0, 0)

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

D (x, y) = 36 x - 9

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, \frac{1}{4}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{2}, \frac{1}{4}), Saddle (0, 0)

What is the extreme f(x,y)=x^3-3xy+3y^2 ?
The extreme f(x,y)=x^3-3xy+3y^2 is Minimum(1/2 , 1/4),Saddle(0,0)