What is the extreme f(x,y)=12-x^2-y^2 ?

The extreme f(x,y)=12-x^2-y^2 is Maximum(0,0)

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extreme f(x,y)=12-x^2-y^2

extreme points

f (x, y) = 12 - 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) = x + y + x y

f (x) = 1 + \frac{5}{x} - \frac{8}{x ^{2}}

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

y = 3 x - x^{3}

f (x, y) = x y - 4 x - 4 y - x^{2} - y^{2}

What is the extreme f(x,y)=12-x^2-y^2 ?
The extreme f(x,y)=12-x^2-y^2 is Maximum(0,0)