  Let $\mathcal{R}$ be a family of axis-parallel rectangles in the plane. The
transversal number $\tau(\mathcal{R})$ is the minimum number of points needed
to pierce all the rectangles. The independence number $\nu(\mathcal{R})$ is the
maximum number of pairwise disjoint rectangles. Given a positive real number
$r$, we say that $\mathcal{R}$ is an r-bounded family if, for any rectangle in
$\mathcal{R}$, the aspect ratio of the longer side over the shorter side is at
most $r$. Gy\'arf\'as and Lehel asked if it is possible to bound the
transversal number $\tau(\mathcal{R})$ with a linear function of the
independence number $\nu(\mathcal{R})$. Ahlswede and Karapetyan claimed a
positive answer for the particular case of $r$-bounded families, but without
providing proof. Chudnovsky et al. confirmed the result proving the bound $\tau
\leq (14 + 2r^2) \nu$. This note aims at giving a simple proof of $\tau \leq
2(r+1)(\nu-1) + 1$, slightly improving the previous results. As a consequence
of this new approach, we also deduce a constant factor bound for the ratio
$\frac{\chi}{\omega}$ in the case of $r$-bounded family.
