What is the extreme f(x)=sqrt(25-x^2-y^2) ?

The extreme f(x)=sqrt(25-x^2-y^2) is Maximum(0,0)

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extreme f(x)=sqrt(25-x^2-y^2)

extreme points

f (x) = 25 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{25 - x ^{2}}{( 25 - x ^{2} - y ^{2} ) 25 - x ^{2} - y ^{2}}

D (x, y) = \frac{25}{( - x ^{2} - y ^{2} + 25 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

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

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

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

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

What is the extreme f(x)=sqrt(25-x^2-y^2) ?
The extreme f(x)=sqrt(25-x^2-y^2) is Maximum(0,0)