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

The extreme f(x,y)=sqrt(4-x^2-y^2) is Maximum(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)=sqrt(4-x^2-y^2)

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = \frac{4}{( - x ^{2} - y ^{2} + 4 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x, y) = x y

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

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

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

What is the extreme f(x,y)=sqrt(4-x^2-y^2) ?
The extreme f(x,y)=sqrt(4-x^2-y^2) is Maximum(0,0)