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

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

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

extreme points

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

D (x, y) = - \frac{x ^{2} + y ^{2} + 4}{( x ^{2} + y ^{2} - 4 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 4 x^{3} - 9 x

f (x) = \frac{∣ x ∣}{2 - x}

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

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

f (x) = 3 x + 3 y

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