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

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

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

extreme points

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

D (x, y) = \frac{16}{( - y ^{2} + 4 - 4 x ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

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

f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 2

f (x) = (x - 1) (x + 1) (x + 4)

\frac{x + 1}{x ^{2} - 5 x + 6}

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