What is the extreme f(x,y)=sqrt(36-x^2-y^2) ?

The extreme f(x,y)=sqrt(36-x^2-y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(36-x^2-y^2)

extreme points

f (x, y) = 36 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{36 - x ^{2}}{( 36 - x ^{2} - y ^{2} ) 36 - x ^{2} - y ^{2}}

D (x, y) = \frac{36}{( - x ^{2} - y ^{2} + 36 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

x 4 - x^{2}

f (x, y) = 4 x^{2} + 8 x + 4 y^{2} + 4

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

f (x, y) = x^{3} + y^{3} + 3 x^{2} - 18 y^{2} + 81 y + 5

What is the extreme f(x,y)=sqrt(36-x^2-y^2) ?
The extreme f(x,y)=sqrt(36-x^2-y^2) is Maximum(0,0)