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

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

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

extreme points

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

Minimum (- 3, 3)

Solution steps

Find the critical points:

(- 3, 3)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- 3, 3) :

Minimum

Minimum (- 3, 3)

f (x) = - 2 x^{2} - 12 x - 13

f (x) = (x^{2} - 1)^{2}

f (x) = x y

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

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

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