What is the extreme f(x)=sqrt(16-4x^2-y^2) ?

The extreme f(x)=sqrt(16-4x^2-y^2) is Maximum(0,0)

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extreme f(x)=sqrt(16-4x^2-y^2)

extreme points

f (x) = 16 - 4 x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{16 - 4 x ^{2}}{( 16 - 4 x ^{2} - y ^{2} ) 16 - 4 x ^{2} - y ^{2}}

D (x, y) = \frac{64}{( - y ^{2} + 16 - 4 x ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

F (m, a) = ma

f (x) = 10 + 12 x - 3 x^{2} - 2 x^{3}

f (x) = 3 x y + 12 y x^{5} - 3

f (x, y) = 6 x^{2} - 2 x^{3} + 3 y^{2} + 6 x y

What is the extreme f(x)=sqrt(16-4x^2-y^2) ?
The extreme f(x)=sqrt(16-4x^2-y^2) is Maximum(0,0)