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

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

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extreme f(x,y)=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}} = - 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

\frac{1}{ln ( 3 )} (x + 3 ln (x))

f (x, y) = 3 x^{2} + 2 x y - 8 y^{2}

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

f (x) = 4 x e^{- x}

f (x) = \frac{e ^{x}}{x ^{10}}

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