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 points

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

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})

f (x) = \frac{x}{2} - sin (x)

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

f (x) = x^{2} - 4 x + 1, 0 \leq x \leq 3

x - 3

f (x, y) = 3 y^{2} + x^{3} - 6 x y - 9 x

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)