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

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

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

extreme points

f (x) = 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) = - 2 x^{3} - 2 y^{3} - 6 x y + 1

f (x) = 3 e^{x} - x e^{x}

P (x, y) = 3 x^{2} - 8 x y - 3 y^{2}

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

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

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