What is the extreme f(x,y)= 1/(sqrt(16-4x^2-y^2)) ?

The extreme f(x,y)= 1/(sqrt(16-4x^2-y^2)) is Minimum(0,0)

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extreme f(x,y)= 1/(sqrt(16-4x^2-y^2))

extreme points

f (x, y) = \frac{1}{16 - 4 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} + 2 ( - x ^{2} + 4 ) )}{( 16 - 4 x ^{2} - y ^{2} ) ^{\frac{5}{2}}}

D (x, y) = \frac{8 ( ( 16 - 4 x ^{2} - y ^{2} ) ^{\frac{3}{2}} + 12 x ^{2} - 4 x ^{2} + 16 - y ^{2} ) ( y ^{2} + 2 ( - x ^{2} + 4 ) ) ( 16 - 4 x ^{2} - y ^{2} ) ^{5} - 144 x ^{2} y ^{2} ( 16 - 4 x ^{2} - y ^{2} ) ^{\frac{11}{2}}}{( 16 - 4 x ^{2} - y ^{2} ) ^{\frac{21}{2}}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{x}{x ^{3} - 2}

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

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

f (x) = x^{4} - 18 x^{2} + 4

f (x) = 49 x + \frac{16}{x}

What is the extreme f(x,y)= 1/(sqrt(16-4x^2-y^2)) ?
The extreme f(x,y)= 1/(sqrt(16-4x^2-y^2)) is Minimum(0,0)