What is the extreme f(x,y)=x^2+xy+y^2-22y+161 ?

The extreme f(x,y)=x^2+xy+y^2-22y+161 is Minimum(-22/3 , 44/3)

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extreme f(x,y)=x^2+xy+y^2-22y+161

extreme points

f (x, y) = x^{2} + x y + y^{2} - 22 y + 161

Minimum (- \frac{22}{3}, \frac{44}{3})

Solution steps

Find the critical points:

(- \frac{22}{3}, \frac{44}{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{22}{3}, \frac{44}{3}) :

Minimum

Minimum (- \frac{22}{3}, \frac{44}{3})

f (x) = 3 x^{6} - 7 x^{5}

f (x, y) = e^{x^{2} + 3 y^{2} + 15}

g (x) = x^{3} - 2 x^{2} - 4 x + 8

f (x) = x^{3} + 6 x^{2} - 63 x + 5, - 8 \leq x \leq 0

f (x) = 3 x^{2} - 6 x + 24

What is the extreme f(x,y)=x^2+xy+y^2-22y+161 ?
The extreme f(x,y)=x^2+xy+y^2-22y+161 is Minimum(-22/3 , 44/3)