What is the extreme Q(x,y)=(x+3y)(x-3xy+9y) ?

The extreme Q(x,y)=(x+3y)(x-3xy+9y) is

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extreme Q(x,y)=(x+3y)(x-3xy+9y)

extreme

Saddle (0, 0), Saddle (2, - \frac{2}{3}), Saddle (\frac{7 + 13}{3}, \frac{1 + 13}{9}), Minimum (\frac{7 - 13}{3}, - \frac{13 - 1}{9})

Solution steps

Find the critical points:

(0, 0), (2, - \frac{2}{3}), (\frac{7 + 13}{3}, \frac{1 + 13}{9}), (\frac{7 - 13}{3}, - \frac{13 - 1}{9})

\frac{\partial ^{2} f}{\partial x ^{2}} = 3 (- 6 x + 18)

D (x, y) = - 36 x^{2} + 108 x - 108 x y + 108 y - 324 y^{2} - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, - \frac{2}{3}) :

Saddle

Check critical point

(\frac{7 + 13}{3}, \frac{1 + 13}{9}) :

Saddle

Check critical point

(\frac{7 - 13}{3}, - \frac{13 - 1}{9}) :

Minimum

Saddle (0, 0), Saddle (2, - \frac{2}{3}), Saddle (\frac{7 + 13}{3}, \frac{1 + 13}{9}), Minimum (\frac{7 - 13}{3}, - \frac{13 - 1}{9})

What is the extreme Q(x,y)=(x+3y)(x-3xy+9y) ?
The extreme Q(x,y)=(x+3y)(x-3xy+9y) is