What is the extreme f(x,y)=-x^2-y^2+16 ?

The extreme f(x,y)=-x^2-y^2+16 is Maximum(0,0)

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extreme f(x,y)=-x^2-y^2+16

extreme points

f (x, y) = - x^{2} - y^{2} + 16

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) = 3 x^{4} - 294 x^{2} + 5

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

f (x, y) = (x - 1)^{2} + y^{2} + 5

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

f (x) = \frac{2 e ^{x}}{x ^{5}}, [0, \infty]

What is the extreme f(x,y)=-x^2-y^2+16 ?
The extreme f(x,y)=-x^2-y^2+16 is Maximum(0,0)