What is the extreme f(x,y)= 1/3 sqrt(36-9x^2-4y^2) ?

The extreme f(x,y)= 1/3 sqrt(36-9x^2-4y^2) is Maximum(0,0)

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extreme f(x,y)= 1/3 sqrt(36-9x^2-4y^2)

extreme points

f (x, y) = \frac{1}{3} 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{12 ( - x ^{2} + 4 )}{( 36 - 9 x ^{2} - 4 y ^{2} ) 36 - 9 x ^{2} - 4 y ^{2}}

D (x, y) = \frac{144}{( 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)

y + v

y = \frac{4 x}{x ^{2} + 1}

f (x) = - 5 x^{6} ln (x), (0, 4)

f (x) = 8 sin (x) + 8 cos (x)

f (x) = 5 x + 5 sin (x), 0 \leq x \leq 2 π

What is the extreme f(x,y)= 1/3 sqrt(36-9x^2-4y^2) ?
The extreme f(x,y)= 1/3 sqrt(36-9x^2-4y^2) is Maximum(0,0)