What is the extreme f(x)=x^2+y^2-8x+y ?

The extreme f(x)=x^2+y^2-8x+y is Minimum(4,-1/2)

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extreme f(x)=x^2+y^2-8x+y

extreme

Minimum (4, - \frac{1}{2})

Solution steps

Find the critical points:

(4, - \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

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

Minimum

Minimum (4, - \frac{1}{2})

What is the extreme f(x)=x^2+y^2-8x+y ?
The extreme f(x)=x^2+y^2-8x+y is Minimum(4,-1/2)