$2x-y-3=0$
$⇒x=\dfrac{y+3}{2}$
Khi đó:
$B=(\dfrac{y+3}{2})^2-y^2+12$
$=\dfrac{(y+3)^2-4y^2+48}{4}$
$=\dfrac{y^2+6y+9-4y^2+48}{4}$
$=\dfrac{-3y^2+6y-3+60}{4}$
$=\dfrac{-3(y^2-2y+1)+60}{4}$
$=\dfrac{60-3(y-1)^2}{4}$
Mà $3.(y-1)^2≥0$
$⇒60-3.(y-1)^2≤60$
$⇒\dfrac{60-3(y-1)^2}{4}$$≤\dfrac{60}{4}=15$
Hay $B≤15$
Dấu $=$ xảy ra $⇔(y-1)^2=0⇔y-1=0⇔y=1⇔x=2;y=1$
Vậy $MaxB=15$ tại $x=2;y=1$