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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = (6 x - x^{2}) (4 y - y^{2})

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

ln (x) - 3 x + 2

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

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

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