What is the extreme f(x,y)=xy(1+2x-3y) ?

The extreme f(x,y)=xy(1+2x-3y) is Saddle(0, 1/3),Minimum(-1/6 , 1/9)

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extreme f(x,y)=xy(1+2x-3y)

extreme

Saddle (0, \frac{1}{3}), Minimum (- \frac{1}{6}, \frac{1}{9})

Solution steps

Find the critical points:

(0, \frac{1}{3}), (- \frac{1}{6}, \frac{1}{9})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6 x

D (x, y) = - 16 x^{2} + 24 x y - 8 x - 36 y^{2} + 12 y - 1

Check the sign of at each critical point

Check critical point

(0, \frac{1}{3}) :

Saddle

Check critical point

(- \frac{1}{6}, \frac{1}{9}) :

Minimum

Saddle (0, \frac{1}{3}), Minimum (- \frac{1}{6}, \frac{1}{9})

What is the extreme f(x,y)=xy(1+2x-3y) ?
The extreme f(x,y)=xy(1+2x-3y) is Saddle(0, 1/3),Minimum(-1/6 , 1/9)