What is the extreme x^2+xy+y^2+4y ?

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

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extreme x^2+xy+y^2+4y

extreme points

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

x^{2} + 2 y^{2} - x y

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

f (x) = x^{4} - 2 x^{2} + x^{3}, - 4 \leq x \leq - 1

x^{2} + x y + y^{2} + 7 y

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