What is the extreme f(x,y)=(sqrt(16-x^2-4y^2))/2 ?

The extreme f(x,y)=(sqrt(16-x^2-4y^2))/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^2-4y^2))/2

extreme points

f (x, y) = \frac{16 - x ^{2} - 4 y ^{2}}{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{2 ( 16 - x ^{2} )}{( 16 - x ^{2} - 4 y ^{2} ) 16 - x ^{2} - 4 y ^{2}}

D (x, y) = \frac{16}{( - x ^{2} + 16 - 4 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = (x + 3)^{2}

y = - x^{2} + 4 x + 2

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

f (t) = 2 cos (t) + sin (2 t)

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

What is the extreme f(x,y)=(sqrt(16-x^2-4y^2))/2 ?
The extreme f(x,y)=(sqrt(16-x^2-4y^2))/2 is Maximum(0,0)