What is the extreme f(x,y)=x^2+y^2-30 ?

The extreme f(x,y)=x^2+y^2-30 is Minimum(0,0)

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extreme f(x,y)=x^2+y^2-30

extreme points

f (x, y) = x^{2} + y^{2} - 30

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

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

s (t) = - t^{2} + 5 t + 1

f (x) = x^{3} + 3 x^{4}

f (x) = - x + 49

What is the extreme f(x,y)=x^2+y^2-30 ?
The extreme f(x,y)=x^2+y^2-30 is Minimum(0,0)