What is the extreme f(x,y)=2x^2+8xy+y^4 ?

The extreme f(x,y)=2x^2+8xy+y^4 is Saddle(0,0),Minimum(4,-2),Minimum(-4,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=2x^2+8xy+y^4

extreme points

f (x, y) = 2 x^{2} + 8 x y + y^{4}

Saddle (0, 0), Minimum (4, - 2), Minimum (- 4, 2)

Solution steps

Find the critical points:

(0, 0), (4, - 2), (- 4, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2}

D (x, y) = 48 y^{2} - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, - 2) :

Minimum

Check critical point

(- 4, 2) :

Minimum

Saddle (0, 0), Minimum (4, - 2), Minimum (- 4, 2)

What is the extreme f(x,y)=2x^2+8xy+y^4 ?
The extreme f(x,y)=2x^2+8xy+y^4 is Saddle(0,0),Minimum(4,-2),Minimum(-4,2)