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

The extreme f(x,y)=2x^2+y^2 is Minimum(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)=2x^2+y^2

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{( 1 )}{( 3 ) x ^{(3)} - 9}

f (x) = \frac{x ^{3}}{3} - 65 x^{2} + 4000 x - 100

f (x, y) = e^{2 x y}

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

f (x) = \frac{6 x ^{2} - x ^{4}}{9}

What is the extreme f(x,y)=2x^2+y^2 ?
The extreme f(x,y)=2x^2+y^2 is Minimum(0,0)