连接$DF$.

$\because CD=2BD$，$CF=2AF$，
$\therefore \frac{CD}{BD}=\frac{CF}{AF}=2$，
$\therefore DF$∥$AB$，
$\therefore \triangle CDF$∽$\triangle CBA$，
$\therefore \frac{DF}{AB}=\frac{CD}{CB}=\frac{2}{3}$，
$\therefore \frac{DE}{AE}=\frac{DF}{AB}=\frac{2}{3}$，
$\because DF$∥$AB$，
$\therefore S_{\triangle ABF}=S_{\triangle ABD}$，
$\therefore S_{\triangle AEF}=S_{\triangle BDE}$，
$\therefore S_{\triangle AEF}=\frac{2}{5}S_{\triangle ABD}$，
$\because BD=\frac{1}{3}BC=\frac{5}{3}$，
$\therefore$当$AB\bot BD$时，$\triangle ABD$的面积最大，最大值$=\frac{1}{2}\times \frac{5}{3}\times 4=\frac{10}{3}$，
$\therefore \triangle AEF$的面积的最大值$=\frac{2}{5}\times \frac{10}{3}=\frac{4}{3}$，
故答案为：$\frac{4}{3}$.

连接$DF$.

$\because CD=2BD$，$CF=2AF$，
$\therefore \frac{CD}{BD}=\frac{CF}{AF}=2$，
$\therefore DF$∥$AB$，
$\therefore \triangle CDF$∽$\triangle CBA$，
$\therefore \frac{DF}{AB}=\frac{CD}{CB}=\frac{2}{3}$，
$\therefore \frac{DE}{AE}=\frac{DF}{AB}=\frac{2}{3}$，
$\because DF$∥$AB$，
$\therefore S_{\triangle ABF}=S_{\triangle ABD}$，
$\therefore S_{\triangle AEF}=S_{\triangle BDE}$，
$\therefore S_{\triangle AEF}=\frac{2}{5}S_{\triangle ABD}$，
$\because BD=\frac{1}{3}BC=\frac{5}{3}$，
$\therefore$当$AB\bot BD$时，$\triangle ABD$的面积最大，最大值$=\frac{1}{2}\times \frac{5}{3}\times 4=\frac{10}{3}$，
$\therefore \triangle AEF$的面积的最大值$=\frac{2}{5}\times \frac{10}{3}=\frac{4}{3}$，
故答案为：$\frac{4}{3}$.