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

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

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

extreme points

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = x^{3} + 12 x

f (x) = 2 x^{3} - 6 x, 0 \leq x \leq 3

\frac{x ^{4}}{x ^{2} - 1}

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

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

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