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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 5

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = x^{4} + y^{4} + 3 x^{2} y + 2 x

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

16

f (x) = - 5 + y - 10 x

f (x, y) = - 6 x^{2} + 7 x y - 2 y^{2} + x + y

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