What is the extreme f(x,y)=x^2+y^2-6x+y ?

The extreme f(x,y)=x^2+y^2-6x+y is Minimum(3,-1/2)

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extreme f(x,y)=x^2+y^2-6x+y

extreme points

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

Minimum (3, - \frac{1}{2})

Solution steps

Find the critical points:

(3, - \frac{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

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

Minimum

Minimum (3, - \frac{1}{2})

f (x, y) = x^{2} - x y + y^{2} - 2 y + x p (1.1)

2 x^{2} - 8 x y + 10 y^{2} + 10 x - 4 y - 5

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

ln (2 x - 3) - ln (x + 5) + 1

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

What is the extreme f(x,y)=x^2+y^2-6x+y ?
The extreme f(x,y)=x^2+y^2-6x+y is Minimum(3,-1/2)