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)

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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) = x^{3} + 5 x - 4

f (x, y) = 2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

f (x, y) = \frac{1}{x} - \frac{64}{y} + x y

f (x) = x 5 - x

f (x) = cos (3 x) - 2, (0, 2 π)

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)