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

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

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

extreme points

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

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} - 4 x y + 4

2 x^{2} + 4 x + 1

P (r, y) = (r + 2 1 + m^{2}) y

f (x) = - x^{2} + 6 x - 4

f (x) = x^{3} - 12 x + 11

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