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

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

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

extreme points

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

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) = 4 - (x - 5)^{5}

f (x) = - 3 x x + 1

f (x) = - 20 x^{6} - 5 x^{3} + x^{2} - 17

f (x) = \frac{π r ^{2}}{2} + \frac{( 76 - π x - x )}{2} \cdot x

10 x^{2} y - 5 x^{2} - 4 y^{2} - x^{4} - 2 y^{4}

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