What is the extreme f(x,y)=1-x^2-y^2 ?

The extreme f(x,y)=1-x^2-y^2 is Maximum(0,0)

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extreme f(x,y)=1-x^2-y^2

extreme points

f (x, y) = 1 - 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)

x^{3} + y^{3} - 3 x y

f (x) = x^{2} + 10 x + 24

f (x, y) = 2 x^{3} - 6 x + 6 x y^{2}

f (x, y) = \frac{3}{2} x^{2} + x^{3} + 3 y^{2}

(x^{2} - 3 x + 5) e^{- \frac{x}{3}}

What is the extreme f(x,y)=1-x^2-y^2 ?
The extreme f(x,y)=1-x^2-y^2 is Maximum(0,0)