What is the extreme f(x,y)=(500)/((4+x^2+y^2)) ?

The extreme f(x,y)=(500)/((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)=(500)/((4+x^2+y^2))

extreme points

f (x, y) = \frac{500}{( 4 + x ^{2} + y ^{2} )}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{1000 ( - 3 y ^{2} + x ^{2} + 4 )}{( 4 + x ^{2} + y ^{2} ) ^{3}}

D (x, y) = \frac{- 3000000 x ^{4} - 6000000 x ^{2} y ^{2} - 8000000 x ^{2} - 3000000 y ^{4} - 8000000 y ^{2} + 16000000}{( x ^{2} + y ^{2} + 4 ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = \frac{5 + x}{4 - x}

f (x) = 7 x^{2} + x - 3

y = 5 x + \frac{180}{x}

\frac{- 8 x ^{3} + 5 x ^{2} - 1}{2 x ^{2} - 9 x}

f (x) = x - 6 x + 9

What is the extreme f(x,y)=(500)/((4+x^2+y^2)) ?
The extreme f(x,y)=(500)/((4+x^2+y^2)) is Maximum(0,0)