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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = 25 x^{2} + 4 y^{2} + 4

f (x) = - 4 x^{3} + 12 x^{2} + 4 x - 12

f (x) = \frac{4 x - 3}{x ^{2} - 4 x + 4}

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

S (a, d) = \frac{n}{2} [2 a + (n - 1) d]

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