What is the extreme f(x,y)=sqrt(1-(x^2)/4-y^2) ?

The extreme f(x,y)=sqrt(1-(x^2)/4-y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(1-(x^2)/4-y^2)

extreme points

f (x, y) = 1 - \frac{x ^{2}}{4} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

f (x) = x^{3} - 1

f (x, y) = x^{4} - \frac{18 x}{12 y ^{2}} - x y^{3} - x^{2}

y = x^{2} - z^{2}

f (x, y) = x^{2} y^{4} + e^{3 x}

What is the extreme f(x,y)=sqrt(1-(x^2)/4-y^2) ?
The extreme f(x,y)=sqrt(1-(x^2)/4-y^2) is Maximum(0,0)