What is the extreme f(x,y)=sqrt(16-x^2-16y^2) ?

The extreme f(x,y)=sqrt(16-x^2-16y^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(16-x^2-16y^2)

extreme points

f (x, y) = 16 - x^{2} - 16 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{16 ( 16 - x ^{2} )}{( 16 - x ^{2} - 16 y ^{2} ) 16 - x ^{2} - 16 y ^{2}}

D (x, y) = \frac{256}{( - x ^{2} + 16 - 16 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = x^{3} + y^{3} + 9 x^{2} - 3 y^{2} - 8

f (x, y) = 6 x - x^{2} + 2 y^{4} - 16 y^{2} + 5

f (x) = x^{\frac{2}{3}} (20 - x)

y = x^{2} + 6 z + z^{2}

f (x) = (x - 1)^{2} (x - 3)

What is the extreme f(x,y)=sqrt(16-x^2-16y^2) ?
The extreme f(x,y)=sqrt(16-x^2-16y^2) is Maximum(0,0)