What is the extreme f(x,y)=sqrt(9-x^2-9y^2) ?

The extreme f(x,y)=sqrt(9-x^2-9y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(9-x^2-9y^2)

extreme points

f (x, y) = 9 - x^{2} - 9 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{9 ( 9 - x ^{2} )}{( 9 - x ^{2} - 9 y ^{2} ) 9 - x ^{2} - 9 y ^{2}}

D (x, y) = \frac{81}{( - x ^{2} + 9 - 9 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x) = 5 - x^{2}

x^{4} - 5 x^{3}

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, - 1 \leq x \leq 10

f (x) = \frac{8}{3 x ^{2} + 4}

What is the extreme f(x,y)=sqrt(9-x^2-9y^2) ?
The extreme f(x,y)=sqrt(9-x^2-9y^2) is Maximum(0,0)