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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

P (x, y) = 6 x^{2} + 19 x y + 15 y^{2} - 17 y - 11 x + 4

5 sin^{2} (x), 0 \leq x \leq π

f (x) = - x^{3} + 3 x - 11

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

f (x) = - 12 x^{3} + 2 x^{2} - 3

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