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

The extreme f(x,y)=x^2-6xy+y^3-3 is Minimum(18,6),Saddle(0,0)

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^2-6xy+y^3-3

extreme points

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

Minimum (18, 6), Saddle (0, 0)

Solution steps

Find the critical points:

(18, 6), (0, 0)

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

D (x, y) = 12 y - 36

Check the sign of at each critical point

Check critical point

(18, 6) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (18, 6), Saddle (0, 0)

f (x) = (x + 9)^{\frac{2}{3}}

- x^{2} + 10 x - 21

f (x) = \frac{x ^{2}}{x ^{2} + 243}

- 5 x^{2} + 4 x y - y^{2} + 16 x + 10

f (x) = x + y + 1

What is the extreme f(x,y)=x^2-6xy+y^3-3 ?
The extreme f(x,y)=x^2-6xy+y^3-3 is Minimum(18,6),Saddle(0,0)