What is the extreme f(x,y)=(20000)/(3+x^2+y^2) ?

The extreme f(x,y)=(20000)/(3+x^2+y^2) is Maximum(0,0)

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extreme f(x,y)=(20000)/(3+x^2+y^2)

extreme points

f (x, y) = \frac{20000}{3 + x ^{2} + y ^{2}}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{40000 ( - 3 y ^{2} + x ^{2} + 3 )}{( 3 + x ^{2} + y ^{2} ) ^{3}}

D (x, y) = \frac{1600000000 ( - 3 x ^{2} + y ^{2} + 3 ) ( - 3 y ^{2} + x ^{2} + 3 ) - 16000 0 ^{2} x ^{2} y ^{2}}{( x ^{2} + y ^{2} + 3 ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 2 x^{3} - 15 x^{2} + 24 x, 0 \leq x \leq 5

y = \frac{x}{x ^{2} + 64}

f (x) = 3 x^{4} - 324 x, 0 \leq x \leq 4

f (x) = - x^{2} + 1, - 1 \leq x \leq 0

f (x) = x^{5} - x^{3} + 8, - 1 \leq x \leq 1

What is the extreme f(x,y)=(20000)/(3+x^2+y^2) ?
The extreme f(x,y)=(20000)/(3+x^2+y^2) is Maximum(0,0)