What is the extreme f(x,y)=(1/(sqrt(16-x^2-y^2))) ?

The extreme f(x,y)=(1/(sqrt(16-x^2-y^2))) is Minimum(0,0)

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extreme f(x,y)=(1/(sqrt(16-x^2-y^2)))

extreme points

f (x, y) = (\frac{1}{16 - x ^{2} - y ^{2}})

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 y ^{2} - x ^{2} + 16}{( 16 - x ^{2} - y ^{2} ) ^{\frac{5}{2}}}

D (x, y) = \frac{- 2 x ^{4} - 4 x ^{2} y ^{2} + 16 x ^{2} - 2 y ^{4} + 16 y ^{2} + 256}{( - x ^{2} - y ^{2} + 16 ) ^{5}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

6 x

(210 - x) (110 + x) - 24 (210 - x)

f (x, y) = 3 x^{2} - y^{2} + 4

f (x) = 12 x^{2} - 48 x + 20

f (x) = 4 x^{4} - 32 x^{3}

What is the extreme f(x,y)=(1/(sqrt(16-x^2-y^2))) ?
The extreme f(x,y)=(1/(sqrt(16-x^2-y^2))) is Minimum(0,0)