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

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

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

extreme points

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 2 x^{4} - 32 x^{3} + 4

f (x) = 8 x e^{- x}

f (x) = x + sin (x), 0 \leq x \leq 2 π

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

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

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