What is the extreme L(x,y)=x^2D^3+(2x+1)D^2+D+xy ?

The extreme L(x,y)=x^2D^3+(2x+1)D^2+D+xy is Saddle(0,-2D^2)

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extreme L(x,y)=x^2D^3+(2x+1)D^2+D+xy

extreme points

L (x, y) = x^{2} D^{3} + (2 x + 1) D^{2} + D + x y

Saddle (0, - 2 D^{2})

Solution steps

Find the critical points:

(0, - 2 D^{2})

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

D (x, y) = - 1

Check the sign of at each critical point

Check critical point

(0, - 2 D^{2}) :

Saddle

Saddle (0, - 2 D^{2})

f (x) = 3 x^{2} + 8 x - 4

f (x) = 10 x^{2} + 20 x

f (x, y) = 30 y - 20 x - 10

T (x, y) = 50 - 6 x^{2} - 2 y^{2}

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

What is the extreme L(x,y)=x^2D^3+(2x+1)D^2+D+xy ?
The extreme L(x,y)=x^2D^3+(2x+1)D^2+D+xy is Saddle(0,-2D^2)