What is the extreme f(x,y)=x^2-xy+y^2+3 ?

The extreme f(x,y)=x^2-xy+y^2+3 is Minimum(0,0)

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extreme f(x,y)=x^2-xy+y^2+3

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

f (x) = ln (x) - 2 x^{2}

f (x, y) = \frac{1}{2} x y

f (x, y) = - 2 x^{3} - 5 x^{2} + 4 x y - y^{2} + 4 y - 7

f (x, y) = 5 x^{2} + 5 y^{2} + 5 x y - 10 x - 5 y + 18

What is the extreme f(x,y)=x^2-xy+y^2+3 ?
The extreme f(x,y)=x^2-xy+y^2+3 is Minimum(0,0)