What is the extreme f(x)=x^2+y^2+18x-4y ?

The extreme f(x)=x^2+y^2+18x-4y is Minimum(-9,2)

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extreme f(x)=x^2+y^2+18x-4y

extreme points

f (x) = x^{2} + y^{2} + 18 x - 4 y

Minimum (- 9, 2)

Solution steps

Find the critical points:

(- 9, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 9, 2) :

Minimum

Minimum (- 9, 2)

f (x) = x^{3} - 9 x^{2}, - 1 \leq x \leq 7

f (x, y) = x^{2} + y^{2} + 9

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

f (x) = 2 + 9 x + 3 x^{2} - x^{3}, - 2 \leq x \leq 6

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 6 y + 19

What is the extreme f(x)=x^2+y^2+18x-4y ?
The extreme f(x)=x^2+y^2+18x-4y is Minimum(-9,2)