What is the extreme f(x,y)=2x^2+3xy+y^2+0x+5 ?

The extreme f(x,y)=2x^2+3xy+y^2+0x+5 is Saddle(0,0)

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extreme f(x,y)=2x^2+3xy+y^2+0x+5

extreme points

f (x, y) = 2 x^{2} + 3 x y + y^{2} + 0 x + 5

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = y^{2} (x - 1)^{2} + x^{2} (y + 4)^{2}

f (x) = (x - 2)^{2} + 1

2 x^{3} - 3 x^{2} - 12 x + 1

f (x) = 324 x - 72 x^{2} + 4 x^{3}

f (x) = 2 x + 3 y

What is the extreme f(x,y)=2x^2+3xy+y^2+0x+5 ?
The extreme f(x,y)=2x^2+3xy+y^2+0x+5 is Saddle(0,0)