What is the extreme f(x)=sqrt(100-x^2-y^2) ?

The extreme f(x)=sqrt(100-x^2-y^2) is Maximum(0,0)

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extreme f(x)=sqrt(100-x^2-y^2)

extreme points

f (x) = 100 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{100 - x ^{2}}{( 100 - x ^{2} - y ^{2} ) 100 - x ^{2} - y ^{2}}

D (x, y) = \frac{100}{( - x ^{2} - y ^{2} + 100 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x, y) = x^{3} - 3 x y + 3 y^{2} + 1

f (x, y) = 4 x^{2} + 2 y^{2} - 2 x y - 10 y - 2 x

f (x) = (x + 2)^{2}

f (x) = cos^{2} (x) - 2 sin (x)

What is the extreme f(x)=sqrt(100-x^2-y^2) ?
The extreme f(x)=sqrt(100-x^2-y^2) is Maximum(0,0)