What is the extreme f(x,y)=sqrt(49-9x^2-y^2) ?

The extreme f(x,y)=sqrt(49-9x^2-y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(49-9x^2-y^2)

extreme points

f (x, y) = 49 - 9 x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{49 - 9 x ^{2}}{( 49 - 9 x ^{2} - y ^{2} ) 49 - 9 x ^{2} - y ^{2}}

D (x, y) = \frac{441}{( - y ^{2} + 49 - 9 x ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

y = X^{2} (X - 3)

f (x) = (3 x - 2)^{5}

f (x) = x^{3} - 3 x + y^{2} - 4 y

f (x) = (x^{2} + 13)^{\frac{1}{3}}

f (x) = \frac{x}{x - 2}, 4 \leq x \leq 6

What is the extreme f(x,y)=sqrt(49-9x^2-y^2) ?
The extreme f(x,y)=sqrt(49-9x^2-y^2) is Maximum(0,0)