Question

Minimum value of $x^2+2xy+3y^2-6x-2y$ $x,y,ϵR$ is

• A. - 9
• B. - 12
• C. - 11
• D. - 10

Answer 1

The correct option is C

- 11

$\text{Let}t=x^2+2xy+3y^2-6x-2y,x,y,ϵR.\therefore x^2+2(y-3)x+3y^2-2y-t=0\Delta ⩾0\Rightarrow 4(y-3{)}^2-4(3y^2-2y-t)\ge 0\Rightarrow y^2+9-6y-3y^2+2y+t\ge 0\Rightarrow -2y^2-4y+9+t\ge 0\text{i.e.},t\ge 2(y^2+2y+1)-11t\ge 2(y+1{)}^2-11\therefore \text{Min is}-11\text{at}y=-1$