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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

\frac{D}{D x}

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

f (x, y) = x^{2} - 2 x y + \frac{1}{3} y^{3} - 3 y

f (x, y) = 3 x - 4 y + 7

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