What is the extreme f(x,y)=sqrt(49-x^2-y^2) ?

The extreme f(x,y)=sqrt(49-x^2-y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(49-x^2-y^2)

extreme points

f (x, y) = 49 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{49 - x ^{2}}{( 49 - x ^{2} - y ^{2} ) 49 - x ^{2} - y ^{2}}

D (x, y) = \frac{49}{( - x ^{2} - y ^{2} + 49 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = \frac{1}{2} x^{4} - 3 x y + 3 y^{4}

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

3 x^{5} - 5 x^{3}

\frac{2}{x ^{12}} - \frac{1}{x ^{6}}

f (x) = \frac{x ^{2} - 4}{x - 1}

What is the extreme f(x,y)=sqrt(49-x^2-y^2) ?
The extreme f(x,y)=sqrt(49-x^2-y^2) is Maximum(0,0)