What is the extreme f(x,y)=3sqrt(16-x^2-y^2) ?

The extreme f(x,y)=3sqrt(16-x^2-y^2) is Maximum(0,0)

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extreme f(x,y)=3sqrt(16-x^2-y^2)

extreme points

f (x, y) = 3 16 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{3 ( 16 - x ^{2} )}{( 16 - x ^{2} - y ^{2} ) 16 - x ^{2} - y ^{2}}

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

1 + \frac{5}{x} - \frac{9}{x ^{2}}

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

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

x^{3} - x^{2} - 8 x + 8

f (x, y) = 2 x y - 4 x - 3 y

What is the extreme f(x,y)=3sqrt(16-x^2-y^2) ?
The extreme f(x,y)=3sqrt(16-x^2-y^2) is Maximum(0,0)