What is the extreme f(x,y)=(100)/(1+x^2+y^2) ?

The extreme f(x,y)=(100)/(1+x^2+y^2) is Maximum(0,0)

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extreme f(x,y)=(100)/(1+x^2+y^2)

extreme points

f (x, y) = \frac{100}{1 + x ^{2} + y ^{2}}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{200 ( - 3 y ^{2} + x ^{2} + 1 )}{( 1 + x ^{2} + y ^{2} ) ^{3}}

D (x, y) = \frac{- 120000 x ^{4} - 240000 x ^{2} y ^{2} - 80000 x ^{2} - 120000 y ^{4} - 80000 y ^{2} + 40000}{( x ^{2} + y ^{2} + 1 ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = \frac{1}{6} x^{3} - x^{2} + 6 x + 11

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

f (x) = 230 + 8 x^{3} + x^{4}

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

\frac{ln ( 4 x )}{x}

What is the extreme f(x,y)=(100)/(1+x^2+y^2) ?
The extreme f(x,y)=(100)/(1+x^2+y^2) is Maximum(0,0)