What is the extreme f(x,y)= 1/12 sqrt(144-16x^2-9y^2) ?

The extreme f(x,y)= 1/12 sqrt(144-16x^2-9y^2) is Maximum(0,0)

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extreme f(x,y)= 1/12 sqrt(144-16x^2-9y^2)

extreme points

f (x, y) = \frac{1}{12} 144 - 16 x^{2} - 9 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{12 ( - x ^{2} + 9 )}{( 144 - 16 x ^{2} - 9 y ^{2} ) 144 - 16 x ^{2} - 9 y ^{2}}

D (x, y) = \frac{144}{( 144 - 16 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) = 80 x^{2} - \frac{52}{3} x^{3} + x^{4}

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

f (x) = 2 x^{2} - 8 x + 4

- (x + 4)^{5}

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

What is the extreme f(x,y)= 1/12 sqrt(144-16x^2-9y^2) ?
The extreme f(x,y)= 1/12 sqrt(144-16x^2-9y^2) is Maximum(0,0)