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 Minimum(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}

Minimum (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) :

Minimum

Minimum (0, 0)

f (x) = \frac{x + 5}{x - 3}, (- 5, 2)

f (x) = - \frac{1}{3} x^{3} - x^{2} - 12

f (x) = - \frac{24 x}{x ^{2} + 9}

- 2 x^{2} - 12 x - 17

f (x) = 5 x e^{- x}, 0 \leq x \leq 2

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 Minimum(0,0)