What is the extreme x^2+xy+y^2+8y ?

The extreme x^2+xy+y^2+8y is Minimum(8/3 ,-16/3)

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extreme x^2+xy+y^2+8y

extreme points

x^{2} + x y + y^{2} + 8 y

Minimum (\frac{8}{3}, - \frac{16}{3})

Solution steps

Find the critical points:

(\frac{8}{3}, - \frac{16}{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{8}{3}, - \frac{16}{3}) :

Minimum

Minimum (\frac{8}{3}, - \frac{16}{3})

5 - x

3 x^{3} - 3 x^{2} - 3 x + 5, - 1 \leq x \leq 2

f (x) = 8 x - 2 x^{2} + 12 x y - 12 y^{2} + y

f (x) = - 3 x^{3} + 9 x^{2} - 8

f (x) = - 3 x^{3} + 9 x^{2} - 4

What is the extreme x^2+xy+y^2+8y ?
The extreme x^2+xy+y^2+8y is Minimum(8/3 ,-16/3)