What is the extreme f(x,y)=x^2+xy+y^2+7y ?

The extreme f(x,y)=x^2+xy+y^2+7y is Minimum(7/3 ,-14/3)

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extreme f(x,y)=x^2+xy+y^2+7y

extreme points

f (x, y) = x^{2} + x y + y^{2} + 7 y

Minimum (\frac{7}{3}, - \frac{14}{3})

Solution steps

Find the critical points:

(\frac{7}{3}, - \frac{14}{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{7}{3}, - \frac{14}{3}) :

Minimum

Minimum (\frac{7}{3}, - \frac{14}{3})

f (x) = \frac{7 ( x - 1 ) ^{2}}{x + 9}, (- 21, - 9)

\frac{7 x - 2}{3}

f (x) = x^{3} + y^{3} - 21 x y

f (x, y) = x^{3} + y^{3} - 3 x^{2} - 6 y^{2} - 9 x

h (x) = x^{3} - 12 x

What is the extreme f(x,y)=x^2+xy+y^2+7y ?
The extreme f(x,y)=x^2+xy+y^2+7y is Minimum(7/3 ,-14/3)