What is the extreme f(x)=sqrt(36-9x^2-4y^2) ?

The extreme f(x)=sqrt(36-9x^2-4y^2) is Maximum(0,0)

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extreme f(x)=sqrt(36-9x^2-4y^2)

extreme points

f (x) = 36 - 9 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 ( 36 - 9 x ^{2} )}{( 36 - 9 x ^{2} - 4 y ^{2} ) 36 - 9 x ^{2} - 4 y ^{2}}

D (x, y) = \frac{1296}{( 36 - 9 x ^{2} - 4 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x) = 2 x^{2} (1 - x^{2})

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

f (x, y) = 12 x y - x^{3} - 6 y^{2}

x^{2} + 2 x

What is the extreme f(x)=sqrt(36-9x^2-4y^2) ?
The extreme f(x)=sqrt(36-9x^2-4y^2) is Maximum(0,0)