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

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

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

extreme points

f (x, y) = \frac{1}{3} x^{3} - 2 x y + y^{2}

Minimum (2, 2), Saddle (0, 0)

Solution steps

Find the critical points:

(2, 2), (0, 0)

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

D (x, y) = 4 x - 4

Check the sign of at each critical point

Check critical point

(2, 2) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (2, 2), Saddle (0, 0)

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