What is the extreme f(x,y)=x^2+xy+y^2-31y+320 ?

The extreme f(x,y)=x^2+xy+y^2-31y+320 is Minimum(-31/3 , 62/3)

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extreme f(x,y)=x^2+xy+y^2-31y+320

extreme points

f (x, y) = x^{2} + x y + y^{2} - 31 y + 320

Minimum (- \frac{31}{3}, \frac{62}{3})

Solution steps

Find the critical points:

(- \frac{31}{3}, \frac{62}{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{31}{3}, \frac{62}{3}) :

Minimum

Minimum (- \frac{31}{3}, \frac{62}{3})

f (x, y) = 5 x + 3 y

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

f (x) = \frac{( x ^{2} - 7 x + 26 )}{x - 5}

f (x) = x^{3} + 8 x^{2} + 16 x + 8, - 7 \leq x \leq 0

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

What is the extreme f(x,y)=x^2+xy+y^2-31y+320 ?
The extreme f(x,y)=x^2+xy+y^2-31y+320 is Minimum(-31/3 , 62/3)