What is the extreme f(x)=sqrt(4-x^2-y^2) ?

The extreme f(x)=sqrt(4-x^2-y^2) is Maximum(0,0)

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extreme f(x)=sqrt(4-x^2-y^2)

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{4 - x ^{2}}{( 4 - x ^{2} - y ^{2} ) 4 - x ^{2} - y ^{2}}

D (x, y) = \frac{4}{( - x ^{2} - y ^{2} + 4 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

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

f (x) = 4 x^{3} - 39 x^{2} + 90 x + 2

f (x, y) = \frac{1}{x + y}

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

What is the extreme f(x)=sqrt(4-x^2-y^2) ?
The extreme f(x)=sqrt(4-x^2-y^2) is Maximum(0,0)