What is the extreme P(x,y)=x^2+5xy-4y^2 ?

The extreme P(x,y)=x^2+5xy-4y^2 is Saddle(0,0)

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extreme P(x,y)=x^2+5xy-4y^2

extreme points

P (x, y) = x^{2} + 5 x y - 4 y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = - 41

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

What is the extreme P(x,y)=x^2+5xy-4y^2 ?
The extreme P(x,y)=x^2+5xy-4y^2 is Saddle(0,0)