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

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

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extreme f(x,y)=x^2+y^2-x-y+1

extreme points

f (x, y) = x^{2} + y^{2} - x - y + 1

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

Solution steps

Find the critical points:

(\frac{1}{2}, \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

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

Minimum

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

f (x, y) = (x - y) (16 - x y)

f (x, y) = (1 + x y) (x + y)

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 1

f (x) = x^{3} - 3 x^{2} - 24 x + 2

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

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