What is the extreme f(xy)=x^3+y^2-6xy+9x+5y+2 ?

The extreme f(xy)=x^3+y^2-6xy+9x+5y+2 is Minimum(4, 19/2),Saddle(2, 7/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(xy)=x^3+y^2-6xy+9x+5y+2

extreme

Minimum (4, \frac{19}{2}), Saddle (2, \frac{7}{2})

Solution steps

Find the critical points:

(4, \frac{19}{2}), (2, \frac{7}{2})

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

D (x, y) = 12 x - 36

Check the sign of at each critical point

Check critical point

(4, \frac{19}{2}) :

Minimum

Check critical point

(2, \frac{7}{2}) :

Saddle

Minimum (4, \frac{19}{2}), Saddle (2, \frac{7}{2})

What is the extreme f(xy)=x^3+y^2-6xy+9x+5y+2 ?
The extreme f(xy)=x^3+y^2-6xy+9x+5y+2 is Minimum(4, 19/2),Saddle(2, 7/2)