What is the extreme f(x,y)=x^2+xy+y^2+8y ?

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

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extreme f(x,y)=x^2+xy+y^2+8y

extreme points

f (x, y) = 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})

f (x, y) = x^{2} + x y + y^{2} + 5 y

\frac{x ^{2} - 9}{x ^{2} - 4}

y = 2 x^{3} - 6 x + 1

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

f (x) = \frac{- 1}{x - 6}

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