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

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

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

extreme points

f (x) = 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) = x + \frac{4}{x}, 0.2 \leq x \leq 8

f (x) = (5 - x)^{\frac{2}{3}} (1 + x)^{\frac{1}{3}}

x^{\frac{1}{3}} (x^{2} - 9)

f (x) = x \cdot e^{x}

f (x) = x^{3} + 2 x^{2} - 20 x, 0 \leq x \leq 3

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