What is the extreme f(x,y)=(x-y)(2x+y)(x+1) ?

The extreme f(x,y)=(x-y)(2x+y)(x+1) is Saddle(0,0),Maximum(-2/3 , 1/3)

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extreme f(x,y)=(x-y)(2x+y)(x+1)

extreme

Saddle (0, 0), Maximum (- \frac{2}{3}, \frac{1}{3})

Solution steps

Find the critical points:

(0, 0), (- \frac{2}{3}, \frac{1}{3})

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

D (x, y) = - 28 x^{2} - 4 x y - 36 x - 4 y^{2} - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{2}{3}, \frac{1}{3}) :

Maximum

Saddle (0, 0), Maximum (- \frac{2}{3}, \frac{1}{3})

What is the extreme f(x,y)=(x-y)(2x+y)(x+1) ?
The extreme f(x,y)=(x-y)(2x+y)(x+1) is Saddle(0,0),Maximum(-2/3 , 1/3)