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 points

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

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})

f (x) = \frac{x}{9 + x ^{2}}

f (x, y) = x^{2} + y^{3} - 6 x y + 3 x + 6 y - 2

f (x) = 2 x^{3} + 9 x^{2} - 24 x

f (x, y) = 2 x^{2} - 4 x + y^{2} - 8 y + 1

f (x) = \frac{x ^{3}}{3} - 6 x^{2}

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)