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

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

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

extreme points

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

Minimum (- 4, 4)

Solution steps

Find the critical points:

(- 4, 4)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- 4, 4) :

Minimum

Minimum (- 4, 4)

f (x, y) = y ln (x + 2 y)

f (x) = 2 x^{3} - 24 x^{2} + 72 x + 7, (3, 7)

x^{4} + y^{4} - 4 x y + x

y = x^{3} + 3 x^{2} + 1

f (x) = - 4 x^{3} - 3 x^{2} + 6 x + 1

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