Let $A$ and $B$ be $C^{\ast}$-algebras and $A \otimes B$ denotes algebraic tensor product of $A$ and $B$. Given representations $\pi_1$ and $\pi_2$ of $A$ and $B$ respectively satisfying $\pi_1(a)\pi_2(b)= \pi_2(b)\pi_1(a)$, we define $\pi_1.\pi_2: A\otimes B \to B(H)$ as $$\pi_1.\pi_2\big(\sum_i a_i \otimes b_i\big)= \sum_i \pi_1(a_i)\pi_2(b_i)$$
Let $x\in A\otimes B$, does there exists representations $\pi_1$ and $\pi_2$ of $A$ and $B$ respectively satisfying $\pi_1(a)\pi_2(b)= \pi_2(b)\pi_1(a)$ and $\|x\|_\text{max}=\|\pi_1.\pi_2(x)\|$
First assume that $A$ and $B$ are unital. You can check that $\lVert\cdot\rVert_\max$ is a (usually incomplete) $C^\ast$-norm on $A\otimes B$. Thus the completion $A\otimes_\max B$ is a $C^\ast$-algebra, which can be realized as $C^\ast$-subalgebra of $B(H)$ for some Hilbert space $H$. This gives a unital $\ast$-representation $$ \pi\colon A\otimes B\twoheadrightarrow A\otimes_\max B\hookrightarrow B(H). $$ Let $\pi_1(a)=\pi(a\otimes 1)$ and $\pi_2(b)=\pi(1\otimes b)$. Then $\pi_1$ and $\pi_2$ are representations with commuting images and $\pi=\pi_1\otimes \pi_2$. Moreover, $\lVert \pi_1\otimes \pi_2(x)\rVert=\lVert \pi(x)\rVert=\lVert x\rVert_\max$.
If $A$ and $B$ are not necessarily unital, you can embed them into their unitizations to get the same result.