How to solve such a maximal entropy oriented optimization problem? I am considering such a problem:
(1) Given a finite dimensional composite system AB whose initial state is a product state of A and B so that $\rho_{AB}=\rho_A\otimes \rho_B$. 
(2) Assuming AB undergoes a joint unitary operation $U_{AB}$ on AB, and the output of system A is given by
$O_A=Tr_B (U_{AB}\rho_{AB}U_{AB}^{+})$
Question:
What's the initial state $\rho_A$ that will result in an $O_A$ with a maximal Von Neumann entropy (for a given $\rho_B$ and $U_{AB}$)?
It seems that $\rho_A=I/n$ (n is the dimension of system A) might be the answer (at least for some special cases I tried). But I have no idea if it's an universal answer.
This reminds me of the channel capacity problem of classical information theory (some similarities if $U_{AB}$ is regarded as a channel). I am not sure if this is a solved problem in quantum information field. Thanks.
 A: The answer will obviously depend very much on $\rho_B$ and $U_{AB}$. 
Let's first assume that the output state with maximal entropy is the maximally mixed state. 
First case: This might not be possible. In particular, there are channels that will map every state to a pure state. Here is one that you could already have known from my answer to your previous question: As $U_{AB}$ you take the swap operator $U_{AB}:=\sum_{ij} |ij\rangle \langle ji|$ and as $\rho_B$ an arbitrary pure state. Then $O_A=\rho_B$ always. 
Second case: However, large classes of channels are not of this form. In particular, there are a lot of interesting channels that are unital, meaning that the maximally mixed state will map to itself. This is true, because a lot of channels (such as many noise channels or all channels without an environment) are entropy-increasing and clearly, entropy increasing channels have to map the maximally mixed state to itself. It doesn't have to be the only state that gets mapped to the mixed state - it could be many of them or even all. For instance, consider the channel where $\rho_B$ is maximally mixed and $U_{AB}$ is the swap operation.
Third case: There are also channels that are not unital but still map some state to the maximally mixed state. Norbert Schuch in the comments gave a very simple construction: You let $|0\rangle\mapsto I/2$ the maximally mixed state and $|1\rangle\mapsto |1\rangle$. 

However, I misunderstood your question and you just wanted the maximal entropy possible per channel. This quantity is known as the maximum output entropy. Still, you will not be able to say so easily which state will achieve the maximum output entropy, because the third example is supposed to tell you that even in the case where that entropy is maximal, it's not always clear which state achieves it.
You are right to think that this has something to do with channel capacities: If you take a channel, the maximum output entropy is a (trivial) upper bound to the classical channel capacity, i.e. the maximum rate at which a quantum channel can send usual classical information.
Given a quantum channel $T$, this capacity is given by 
$$ C(T)= \limsup_{n\to \infty} \chi(T^{\otimes n})/n $$
where $\chi$ is the Holevo quantity:
$$ \chi(T)=\max_{p_i} S(T(\sum_{i}p_i\rho_i))-\sum_i p_i S(\rho_i) $$
where $\rho_i$ are the possible messages (see Peter Shor's article for an overview).
Clearly, this is upper bounded by the maximum entropy (just leave out the second part of the term and don't take the limit). 
The bound I believe is mostly not very tight and sometimes referred to as "trivial upper bound". In particular, all unital channels will have maximal output entropy given to the maximum possible von Neumann entropy. In these cases, it will not be very useful to know the maximum output entropy in order to say something about the channel capacity without also knowing the minimum output entropy, which is a much more interesting object: it tells you something about how much noise the channel introduces. If this is high, no matter what you do, all output states will be close to maximally mixed, hence the capacity will probably not be very high. 
