What is the extreme f(x,y)=x^2+y^2-2x+5y ?

The extreme f(x,y)=x^2+y^2-2x+5y is Minimum(1,-5/2)

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extreme f(x,y)=x^2+y^2-2x+5y

extreme points

f (x, y) = x^{2} + y^{2} - 2 x + 5 y

Minimum (1, - \frac{5}{2})

Solution steps

Find the critical points:

(1, - \frac{5}{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, - \frac{5}{2}) :

Minimum

Minimum (1, - \frac{5}{2})

f (x) = 3 - 4 x - \frac{4}{x}

f (x, y) = - 3 x^{2} - 4 x y - 8 y^{2} + 42 x + 68 y + 4

f (x) = \frac{1}{3} x^{3} - x^{2} + 3 x + 4

f (x) = 2 (2 x - 8)

f (x) = - x^{2} - 4 x - 9

What is the extreme f(x,y)=x^2+y^2-2x+5y ?
The extreme f(x,y)=x^2+y^2-2x+5y is Minimum(1,-5/2)