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

The extreme f(x,y)=xy+2xy^2 is Saddle(0,0),Saddle(0,-1/2)

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

extreme points

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

Saddle (0, 0), Saddle (0, - \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (0, - \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 x

D (x, y) = - 16 y^{2} - 8 y - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - \frac{1}{2}) :

Saddle

Saddle (0, 0), Saddle (0, - \frac{1}{2})

(x + 4)^{3} (3 x - 2)^{2}

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

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

f (x) = 14 x^{2} - 2 x^{3} 2 y^{2} + 4 x y

2 x^{2} + y^{2} + x y - 3 x

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