What is the extreme f(x,y)=x^2+xy+y^2+6x-3y+4 ?

The extreme f(x,y)=x^2+xy+y^2+6x-3y+4 is Minimum(-5,4)

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extreme f(x,y)=x^2+xy+y^2+6x-3y+4

extreme points

f (x, y) = x^{2} + x y + y^{2} + 6 x - 3 y + 4

Minimum (- 5, 4)

Solution steps

Find the critical points:

(- 5, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- 5, 4) :

Minimum

Minimum (- 5, 4)

x^{3} - 3 x^{2} + 3 x - 1

f (x) = 5 x + 9 x^{- 1}

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

x^{4} + (1 - x)^{4} + 10 x^{2} (1 - x)^{2}

- 7 x + 6

What is the extreme f(x,y)=x^2+xy+y^2+6x-3y+4 ?
The extreme f(x,y)=x^2+xy+y^2+6x-3y+4 is Minimum(-5,4)