What is the extreme f(x,y)=x^3-18xy+y^3 ?

The extreme f(x,y)=x^3-18xy+y^3 is Saddle(0,0),Minimum(6,6)

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^3-18xy+y^3

extreme points

f (x, y) = x^{3} - 18 x y + y^{3}

Saddle (0, 0), Minimum (6, 6)

Solution steps

Find the critical points:

(0, 0), (6, 6)

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

D (x, y) = 36 x y - 324

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(6, 6) :

Minimum

Saddle (0, 0), Minimum (6, 6)

f (x) = x^{4} - 3 x^{3} + x^{2}

f (x) = \frac{4}{x ^{2}}

F (x, y) = x^{3} - 2 x^{2} y^{2} - 2 x - 4 y + 10

(x + 3)^{4} (2 - x)

f (x, y) = x^{2} + x y + y^{2} + 8 y

What is the extreme f(x,y)=x^3-18xy+y^3 ?
The extreme f(x,y)=x^3-18xy+y^3 is Saddle(0,0),Minimum(6,6)