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

The extreme f(x,y)=x^3-3xy+1/2 y^2 is Minimum(3,9),Saddle(0,0)

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

extreme

Minimum (3, 9), Saddle (0, 0)

Solution steps

Find the critical points:

(3, 9), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 1

D (x, y) = 6 x - 9

Check the sign of at each critical point

Check critical point

(3, 9) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (3, 9), Saddle (0, 0)

What is the extreme f(x,y)=x^3-3xy+1/2 y^2 ?
The extreme f(x,y)=x^3-3xy+1/2 y^2 is Minimum(3,9),Saddle(0,0)