What is the extreme f(x)=sqrt(36-x^2-y^2) ?

The extreme f(x)=sqrt(36-x^2-y^2) is Maximum(0,0)

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extreme f(x)=sqrt(36-x^2-y^2)

extreme points

f (x) = 36 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{36 - x ^{2}}{( 36 - x ^{2} - y ^{2} ) 36 - x ^{2} - y ^{2}}

D (x, y) = \frac{36}{( - x ^{2} - y ^{2} + 36 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x, y) = e^{- x y}

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

w (x, y) = 6 (x + y)

f (x) = \frac{- cos ( 2 x )}{2} - 2 sin (x)

What is the extreme f(x)=sqrt(36-x^2-y^2) ?
The extreme f(x)=sqrt(36-x^2-y^2) is Maximum(0,0)