What is the extreme f(x,y)=5004+x^2+y^2 ?

The extreme f(x,y)=5004+x^2+y^2 is Minimum(0,0)

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extreme f(x,y)=5004+x^2+y^2

extreme points

f (x, y) = 5004 + x^{2} + y^{2}

Minimum (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) :

Minimum

Minimum (0, 0)

f (x) = 2 x^{3} - 6 x^{2} - 210 x + 1, - 6 \leq x \leq 8

f (x) = ln (x^{2} + 3 x + 9)

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

f (x) = \frac{( x - ln ( x ) )}{x}

f (x, y) = e^{10 x^{2} + 4 y^{2} + 2}

What is the extreme f(x,y)=5004+x^2+y^2 ?
The extreme f(x,y)=5004+x^2+y^2 is Minimum(0,0)