What is the extreme f(x,y)=sqrt(11-x^2-y^2) ?

The extreme f(x,y)=sqrt(11-x^2-y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(11-x^2-y^2)

extreme points

f (x, y) = 11 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{11 - x ^{2}}{( 11 - x ^{2} - y ^{2} ) 11 - x ^{2} - y ^{2}}

D (x, y) = \frac{11}{( - x ^{2} - y ^{2} + 11 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = x^{2} - 25

U (X, Y) = X^{0.3} Y^{0.7}

f (x) = x^{2} + \frac{360}{x}

f (x) = x^{3} - 6 x^{2} + 5, (- 3, 5)

f (x) = 0.25 x^{4} + x^{3}

What is the extreme f(x,y)=sqrt(11-x^2-y^2) ?
The extreme f(x,y)=sqrt(11-x^2-y^2) is Maximum(0,0)