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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Saddle (0, 0), Minimum (\frac{2}{3}, \frac{2}{3})

Solution steps

Find the critical points:

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

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

D (x, y) = 36 x y - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Minimum

Saddle (0, 0), Minimum (\frac{2}{3}, \frac{2}{3})

x^{2} - y^{2}

f (x, y) = ln (x + y + 1)

K (r, s) = 4 r - 7 s

f (x) = \frac{x}{81 + x ^{2}}

f (x) = \frac{x ^{2} - 5}{x - 3}

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