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

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

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

extreme points

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

Minimum (1, - 2)

Solution steps

Find the critical points:

(1, - 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, - 2) :

Minimum

Minimum (1, - 2)

r - s - 1

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

f (x) = - 2 x^{2} + 64 x - 80

f (x, y) = x^{2} + y^{2} + 6 x + 10

f (x) = - 34.3 x^{2} + 460.2 x

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