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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = - 17

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{2} y + x y^{2} + 4 x y - 5 y + 3

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

f (x) = \frac{4}{( x - 3 ) ^{2}}

f (x) = 11 + 7 x + x^{2}

f (x) = \frac{3 x}{x ^{2} + 7}, 0 \leq x \leq 4

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