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) = e^{x} (15 - x^{2})

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

f (x) = 10 + 4 x + \frac{9}{x}

f (x) = 8 x^{3} - 17 x^{2} + 12 x + 14

f (x, y) = 3 y^{3} + 7 x y - 9 x^{2} - 5

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)