What is the extreme f(x)=x^2+xy+y^2-19y+120 ?

The extreme f(x)=x^2+xy+y^2-19y+120 is Minimum(-19/3 , 38/3)

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extreme f(x)=x^2+xy+y^2-19y+120

extreme points

f (x) = x^{2} + x y + y^{2} - 19 y + 120

Minimum (- \frac{19}{3}, \frac{38}{3})

Solution steps

Find the critical points:

(- \frac{19}{3}, \frac{38}{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{19}{3}, \frac{38}{3}) :

Minimum

Minimum (- \frac{19}{3}, \frac{38}{3})

f (x) = x^{2} + x - 12

f (x) = \frac{x + 4}{x ^{2} - x - 20}

f (x) = - 3 x^{2} ln (4 x)

f (x) = 3 x^{2} - 6 x + 8

f (x) = \frac{1}{x} - \frac{2}{x ^{2}}, - 2 \leq x \leq 1

What is the extreme f(x)=x^2+xy+y^2-19y+120 ?
The extreme f(x)=x^2+xy+y^2-19y+120 is Minimum(-19/3 , 38/3)