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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{3 ( 36 - 4 x ^{2} )}{( 36 - 4 x ^{2} - 9 y ^{2} ) 36 - 4 x ^{2} - 9 y ^{2}}

D (x, y) = \frac{144}{( 36 - 4 x ^{2} - 9 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 0.2 (x^{4} + 4 x^{3} - 16 x - 16) + 5

f (x) = \frac{3 x}{x ^{2} + 2 x - 8}

f (x) = - x^{7} ln (3 x), (0, 4)

\frac{x ^{2}}{- x ^{2} + 4}

y = x^{x} I n (x^{5} + x)

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