What is the extreme f(x,y)=sqrt(64-4x^2-y^2) ?

The extreme f(x,y)=sqrt(64-4x^2-y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(64-4x^2-y^2)

extreme points

f (x, y) = 64 - 4 x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{64 - 4 x ^{2}}{( 64 - 4 x ^{2} - y ^{2} ) 64 - 4 x ^{2} - y ^{2}}

D (x, y) = \frac{256}{( - y ^{2} + 64 - 4 x ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

\frac{- 8 x + 75}{9 x - 61}

4 + (7 + 5 x)^{\frac{2}{5}}

2.8 x

f (x) = - \frac{x ^{2} - 50}{x ^{2} - 25}

What is the extreme f(x,y)=sqrt(64-4x^2-y^2) ?
The extreme f(x,y)=sqrt(64-4x^2-y^2) is Maximum(0,0)