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}} = - 2

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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)