What is the extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 ?

The extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 is Minimum(-7,-2)

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extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4

extreme

Minimum (- 7, - 2)

Solution steps

Find the critical points:

(- 7, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(- 7, - 2) :

Minimum

Minimum (- 7, - 2)

What is the extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 ?
The extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 is Minimum(-7,-2)