What is the extreme f(x,y)=x^2+xy+y^2-7y+16 ?

The extreme f(x,y)=x^2+xy+y^2-7y+16 is Minimum(-7/3 , 14/3)

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extreme f(x,y)=x^2+xy+y^2-7y+16

extreme points

f (x, y) = x^{2} + x y + y^{2} - 7 y + 16

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, y) = x^{2} + x y + y^{2} - y

f (x) = 3 x^{5} - 5 x^{3} + 2

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

f (x) = x - 1

What is the extreme f(x,y)=x^2+xy+y^2-7y+16 ?
The extreme f(x,y)=x^2+xy+y^2-7y+16 is Minimum(-7/3 , 14/3)