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

The extreme f(x,y)=x^2+y^2-4x+2y is Minimum(2,-1)

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

extreme points

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

Minimum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(2, - 1) :

Minimum

Minimum (2, - 1)

f (x) = \frac{- x ^{2} - 10 x - 13}{x + 9}

x^{3} - 9 x^{2} + 1

f (x) = \frac{- 24 x ^{2} - 20 x + 24}{( x ^{2} + 1 ) ^{2}}

f (x) = x^{2} + y^{2} + 20 x - 8 y

f (x) = x + y + \frac{8}{x y}

What is the extreme f(x,y)=x^2+y^2-4x+2y ?
The extreme f(x,y)=x^2+y^2-4x+2y is Minimum(2,-1)