I would like to solve the following problem for a general $n$:
Find disjoint sets $P$ and $Q$ of prime numbers so that \begin{align*} \pi_P = \prod_{p \in P} p \leq n \textrm{ and } \pi_Q = \prod_{q \in Q} q \leq n \end{align*} and the expression \begin{align*} B(P, Q) &= \Big(\Big\lfloor \frac{n}{\pi_P} \Big\rfloor + 1\Big)\prod_{p \in P} p - 1 + \Big(\Big\lfloor \frac{n}{\pi_Q} \Big\rfloor + 1\Big)\prod_{q \in Q} q - 1 \\ &= \Big(\Big\lfloor \frac{n}{\pi_P} \Big\rfloor + 1\Big)\varphi(\pi_P) + \Big(\Big\lfloor \frac{n}{\pi_Q} \Big\rfloor + 1\Big)\varphi(\pi_Q) \end{align*} is minimised.
I am also interested in any heuristically good solutions where there is no proof of minimality. This problem comes from the bounding the Cheeger constant for the graph with vertex set $[n]$, in which two vertices are adjacent if and only if they are relatively prime.
I should perhaps demonstrate that I have made some attempt on this problem. We can continue bounding above to obtain \begin{align} \Big(\Big\lfloor \frac{n}{\pi_P} \Big\rfloor + 1\Big)\varphi(\pi_P) + \Big(\Big\lfloor \frac{n}{\pi_Q} \Big\rfloor + 1\Big)\varphi(\pi_Q) \leq \Big(\frac{n}{\pi_P} + 1\Big)\pi_P + \Big(\frac{n}{\pi_Q} + 1\Big)\pi_Q = 2n + \pi_P + \pi_Q \end{align} and this suggests that we should take $P = \{2\}$ and $Q = \{3\}$. This gives $B(\{2\}, \{3\}) = \lfloor n/2 \rfloor + 2\lfloor n/3 \rfloor + 3$, but much information is lost when we bound as we just have and so I do not think that this solution is minimal.