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

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

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extreme f(x,y)=x^2+4xy+y^2

extreme points

f (x, y) = x^{2} + 4 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) = - 12

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{3} - 18 x^{2} + 96 x

f (x) = x^{\frac{1}{3}} (x^{2} - 9)

f (x, y) = 3 x^{2} + 2 y^{2} - x y - 4 x - 7 y + 12

\frac{x ^{3} - 4}{x ^{2}}

f (x) = - x^{3} + 12 x

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