What is the extreme f(x,y)=x^2+xy+y^2+4y ?

The extreme f(x,y)=x^2+xy+y^2+4y is Minimum(4/3 ,-8/3)

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extreme f(x,y)=x^2+xy+y^2+4y

extreme

Minimum (\frac{4}{3}, - \frac{8}{3})

Solution steps

Find the critical points:

(\frac{4}{3}, - \frac{8}{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{4}{3}, - \frac{8}{3}) :

Minimum

Minimum (\frac{4}{3}, - \frac{8}{3})

What is the extreme f(x,y)=x^2+xy+y^2+4y ?
The extreme f(x,y)=x^2+xy+y^2+4y is Minimum(4/3 ,-8/3)