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

The extreme f(x,y)=x^2+y^2-xy+2x+y is Minimum(-5/3 ,-4/3)

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

extreme points

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

Minimum (- \frac{5}{3}, - \frac{4}{3})

Solution steps

Find the critical points:

(- \frac{5}{3}, - \frac{4}{3})

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- \frac{5}{3}, - \frac{4}{3}) :

Minimum

Minimum (- \frac{5}{3}, - \frac{4}{3})

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

E (x, y) = 4 x^{2} - 6 x + 9 + y - 8 y^{2} + 5

f (x) = - 3 x^{4} + 6 x^{2} + 1

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

f (x) = 4 cos (x)

What is the extreme f(x,y)=x^2+y^2-xy+2x+y ?
The extreme f(x,y)=x^2+y^2-xy+2x+y is Minimum(-5/3 ,-4/3)