What is the extreme f(x,y)=x^3+y^3-36xy ?

The extreme f(x,y)=x^3+y^3-36xy is Saddle(0,0),Minimum(12,12)

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^3-36xy

extreme points

f (x, y) = x^{3} + y^{3} - 36 x y

Saddle (0, 0), Minimum (12, 12)

Solution steps

Find the critical points:

(0, 0), (12, 12)

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

D (x, y) = 36 x y - 1296

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(12, 12) :

Minimum

Saddle (0, 0), Minimum (12, 12)

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

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

f (x, y) = 2 x^{2} - 4 x + y^{2} - 4 y + 1

3 x^{2} - 24 ln (x)

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

What is the extreme f(x,y)=x^3+y^3-36xy ?
The extreme f(x,y)=x^3+y^3-36xy is Saddle(0,0),Minimum(12,12)