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

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

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

extreme

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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