The binary product of two presentable objects The binary product of two $\lambda$-presentable objects (in a locally presentable category) is $\mu$-presentable for some regular cardinal $\mu \geq \lambda$ (because all objects are $\mu$-presentable for some regular cardinal $\mu$). I don't see any reason for $\mu$ to be equal to $\lambda$. Even if $\lambda^2=\lambda$ and even if the binary product of two finite sets is finite for example.

Is there one ? Or is there a regular cardinal big enough $\mu$ such
  that the class of $\mu$-presentable objects is closed under binary
  product ?

 A: Unfortunately, no example for the first question is coming to mind at the moment.
For the second question, let $\lambda$ be such that the product functor $\times: \mathcal C \times \mathcal C \to \mathcal C$ is $\lambda$-accessible [1]. Let $\mu \rhd \lambda$ be such that the binary product of $\lambda$-presentable objects is $\mu$-presentable. I claim that $\mu$-presentable objects are closed under binary products.
To see this, let $A,B$ be $\mu$-presentable. Then $(A,B)$ is $\mu$-presentable in $\mathcal C \times \mathcal C$ [2]. Write $(A,B) = \varinjlim_{i \in I} (A_i, B_i)$ where $I$ is $\lambda$-filtered and $\mu$-small [3]. Because $\times$ is $\lambda$-accessible, we have 
$$A \times B = \varinjlim_I A_i \times B_i \qquad (\ast)$$
If $C = \varinjlim_{j \in J} C_j$ is a $\mu$-filtered colimit, then we have
$$\mathcal C(A \times B, C) = \mathcal C(\varinjlim_{i \in I} A_i \times B_i, \varinjlim_{j \in J} C_j) = \varprojlim_{i \in I} \mathcal C(A_i \times B_i, \varinjlim_{j \in J} C_j) \\
= \varprojlim_{i \in I} \varinjlim_{j \in J} \mathcal C(A_i \times B_i, C_j) \\
= \varinjlim_{j \in J} \varprojlim_{i \in I} \mathcal C(A_i \times B_i, C_j) \\
= \varinjlim_{j \in J} \mathcal C(\varinjlim_{i \in I} A_i \times B_i, C_j) 
 = \varinjlim_{j \in J} \mathcal C(A \times B, C_j) $$
as desired. In the first line, we have used $(\ast)$. In the second line, we have used that $A_i \times B_i$ is $\mu$-presentable and $J$ is $\mu$-filtered. In the third line we have used that $I$ is $\mu$-small and $J$ is $\mu$-filtered. In the fourth line, we have used $(\ast)$ again.

[1] In fact any $\lambda$ will do here: recall that a right adjoint functor is $\lambda$-accessible iff its left adjoint preserves $\lambda$-presentable objects. The left adjoint of $\times$ is the diagonal functor $\Delta: \mathcal C \to \mathcal C \times \mathcal C$, which preserves $\lambda$-presentable objects for every $\lambda$.
[2] To see this, suppose that $A = \varinjlim_{k \in K} A_k$ and $B = \varinjlim_{l \in L} B_l$ where $K,L$ are $\mu$-small and $\lambda$-filtered. Then we claim that $(A,B) = \varinjlim_{(k,l) \in K \times L} (A_k, B_l)$; then we can take $I = K \times L$, which is also $\mu$-small and $\lambda$-filtered. It suffices to show that the projection $\pi: K \times L \to K$ is cofinal (and dually). Now, for $k \in K$, $k \downarrow \pi = (k \downarrow K ) \times L$, which is a product of filtered categories and so filtered itself, and in particular nonempty and connected. So indeed $\pi$ is cofinal.
[3] Thanks to Mike Shulman, who points out below that this is Rmk 2.15 in Adámek and Rosický -- and rather, $A \times B$ is a retract of such a colimit -- but this is sufficient for our purposes.
