What is the extreme f(x,y)=sqrt(25-x^2-y^2) ?

The extreme f(x,y)=sqrt(25-x^2-y^2) is Maximum(0,0)

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extreme f(x,y)=sqrt(25-x^2-y^2)

extreme points

f (x, y) = 25 - x^{2} - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{25 - x ^{2}}{( 25 - x ^{2} - y ^{2} ) 25 - x ^{2} - y ^{2}}

D (x, y) = \frac{25}{( - x ^{2} - y ^{2} + 25 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = (\frac{x + y - x y}{x - y + x y})

3 (x + 2) (x - 10)

f (x, y) = 1 - x^{2} - x - 3 y^{2} + y

f (x) = 1.4 t e^{- 2.9 t}

f (x) = 2 x^{3} - 9 x^{2} - 24 x + 9, - 2 \leq x \leq 5

What is the extreme f(x,y)=sqrt(25-x^2-y^2) ?
The extreme f(x,y)=sqrt(25-x^2-y^2) is Maximum(0,0)