What is the extreme f(x,y)=3-x^2-y^2 ?

The extreme f(x,y)=3-x^2-y^2 is Maximum(0,0)

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extreme f(x,y)=3-x^2-y^2

extreme points

f (x, y) = 3 - 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)

f (x, y) = 2 x y - 5 x^{2} - 2 y^{2} + 18 y - 8

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

f (x) = \frac{1}{4} x^{4} + \frac{4}{3} x^{3} + 2 x^{2}

f (x) = (1 - x^{2})^{2}

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

What is the extreme f(x,y)=3-x^2-y^2 ?
The extreme f(x,y)=3-x^2-y^2 is Maximum(0,0)