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

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

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

extreme points

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x) = - x^{3} - 7 x^{2} - 16 x - 42

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

x + \frac{64}{x}

f (x) = x^{3} - 6 x^{2} - 15 x + 10

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