Minimum $\mathcal{F}$-covers: the supersolvable and metabelian cases

Given a set $\mathcal{F}$ of finite groups, it is said that a group $G$ is an $\mathcal{F}$-cover if every group in $\mathcal{F}$ is isomorphic to a subgroup of $G$. Moreover, $G$ is a minimum $\mathcal{F}$-cover if there is no $\mathcal{F}$-cover whose order is less than $\mathrm{\mid }G\mathrm{\mid }$. In [Cameron P. J., et al., Minimal cover groups, J. Algebra 660 (2024)], the authors pose the following question: For which classes $\mathcal{X}$ of groups, closed under taking subgroups and direct products, is it true that, if $\mathcal{F}$ is a set of $\mathcal{X}$-groups, then there is a minimum $\mathcal{F}$-cover which is an $\mathcal{X}$-group? In this paper, we give a negative answer in two cases: $\mathcal{X}\in \{\mathrm{‘}\mathrm{‘}supersolvable\mathrm{"},\mathrm{‘}\mathrm{‘}metabelian\mathrm{"}\}\mathrm{.}$