What is the extreme f(x,y)=x^3-9xy-4y^2 ?

The extreme f(x,y)=x^3-9xy-4y^2 is Maximum(-27/8 , 243/64),Saddle(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=x^3-9xy-4y^2

extreme points

f (x, y) = x^{3} - 9 x y - 4 y^{2}

Maximum (- \frac{27}{8}, \frac{243}{64}), Saddle (0, 0)

Solution steps

Find the critical points:

(- \frac{27}{8}, \frac{243}{64}), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = - 48 x - 81

Check the sign of at each critical point

Check critical point

(- \frac{27}{8}, \frac{243}{64}) :

Maximum

Check critical point

(0, 0) :

Saddle

Maximum (- \frac{27}{8}, \frac{243}{64}), Saddle (0, 0)

y = 4 - 6 x^{2}

\frac{x ^{2} + 2}{x - 3}

f (x) = 5 + \frac{48}{x} + 3 x^{2}

E (x, y) = 15 x y - 22 x y

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

What is the extreme f(x,y)=x^3-9xy-4y^2 ?
The extreme f(x,y)=x^3-9xy-4y^2 is Maximum(-27/8 , 243/64),Saddle(0,0)