`~Chris~`
$B=4x-x^2-9y^2-6y-5$
$B=(-9x^2-6y-1)+(-x^2+4x-4)$
$B=-(9y^2+6y+1)-(x^2-4x+4)$
$B=-(3y+1)^2-(x-2)^2$
Vì $-(3y+1)^2 \le 0 \;\forall x \in \mathbb{R}$
$⇒ -(3y+1)^2-(x-2)^2 \le 0\;\forall x\in \mathbb{R}$
Vậy$ \max B=0$ khi $\begin{cases}y=-\dfrac13\\x=2\end{cases}$