Let $P$ be a polyhedron, i.e. $P = \{ x \in \mathbb{R}^{n}\, |\,\, a_{i}^{T}x \leq b_{i} \}$. Define $R$ as the rectangle given by $\{ x \in \mathbb{R}^{n}\, \mid\, \, l \preceq x \preceq u \}$.
Find $l$ and $u$ which maximizes the volume of the rectangle such that $R \subset P$. Formulate the problem as a convex optimization problem without using an exponential number of constraints
My try:
Represent $R$ as the set $S = \{x_{c} + u \, \mid \, \|u_1\| \leq r_1 ... \|u_n\| \leq r_n \}$, where $x_c$ is the center of the rectangle and $r_{i} = \frac{u_i - l_i}{2}$ and each $u_{i} \in \mathbb{R}$.
For one half-space we get the constraint
$$a^{T}x_c + a^{T}u \leq b.$$
To get the biggest value on $u$ we consider
$$\sup\{ a_1u_1 + ... a_nu_n \mid \|u_i \| \leq r_i \forall i = 1...n \},$$
clearly we can maximize each term individually and conclude that $u_{i} = \frac{a_i}{\|a_{i} \|}r_{i}, a_i \neq 0$ and $u_i = 0 $ if $a_i = 0$.
By defining
$$\tilde{a} = diag(\tilde{a}_1,...,\tilde{a}_{n})a,)$$ where $$ \tilde{a}_{i} = \frac{a_i}{\|a_{i}\|}, a_i \neq 0 $$ $$ \tilde{a}_i = 0, a_i = 0$$
we get the constraint
$$a^{T}x_c + \tilde{a}^{T}r \leq b.$$
The same way we can impose the entire polyhedron by the constraints
$$a_1^{T}x_c + \tilde{a}_1^{T}r \leq b_1 \\ \quad \vdots \\ a^{T}_nx_c + \tilde{a}_n^{T}r \leq b_n$$
and end up with the problem
$$\max \prod_{i=1}^{n}2r_{i}$$
$$ \textit{subject to: }a_1^{T}x_c + \tilde{a}_1^{T}r \leq b_1 \\ \quad \vdots \\ a^{T}_nx_c + \tilde{a}_n^{T}r \leq b_n$$
$$ r \succ 0 $$
To get a convex function we maximize $\log(\prod_{i=1}^{n} r_i)$, i.e.
$$\max_{x_c,r} \sum_{i=1}^{n} \log(r_i)$$
$$ \textit{subject to: }a_1^{T}x_c + \tilde{a}_1^{T}r \leq b_1 \\ \quad \vdots \\ a^{T}_nx_c + \tilde{a}_n^{T}r \leq b_n$$
$$ r \succ 0. $$
With this I think all constraints are fulfilled and with convex constraints and concave objective function we have a concave maximization problem, which is equivalent to a convex minimization problem.
Have I missed something or made any mistakes?