What is the extreme f(x,y)=8x^3+2xy-3x^2+y^2+1 ?

The extreme f(x,y)=8x^3+2xy-3x^2+y^2+1 is Minimum(1/3 ,-1/3),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)=8x^3+2xy-3x^2+y^2+1

extreme points

f (x, y) = 8 x^{3} + 2 x y - 3 x^{2} + y^{2} + 1

Minimum (\frac{1}{3}, - \frac{1}{3}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{3}, - \frac{1}{3}), (0, 0)

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

D (x, y) = 96 x - 16

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, - \frac{1}{3}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{3}, - \frac{1}{3}), Saddle (0, 0)

What is the extreme f(x,y)=8x^3+2xy-3x^2+y^2+1 ?
The extreme f(x,y)=8x^3+2xy-3x^2+y^2+1 is Minimum(1/3 ,-1/3),Saddle(0,0)