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

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

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

extreme points

f (x, y) = 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) = x^{4} + y^{4} - 4 x y + 1

f (x) = 3 x^{\frac{2}{3}} - 2 x, - 1 \leq x \leq 1

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

f (x) = x^{3} - 9 x^{2} + 24 x - 7

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

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