= Maximal dimension in mixed SU(3) tensor powers
{title2=$D_{\max}(k,l)$}
For $d(p,q)=(p+1)(q+1)(p+q+2)/2$, the exact largest constituent <dimension> of a <mixed SU(3) tensor representation> is
$$
D_{\max}(k,l)=\max\left\{\max_{0\le a\le\lfloor k/2\rfloor}d(k-2a,l+a),\ \max_{0\le b\le\lfloor l/2\rfloor}d(k+b,l-2b)\right\}.
$$
Every constituent <highest weight> lies below $k\omega_1+l\omega_2$, so its labels have the form $(p,q)=(k-2a+b,l+a-2b)$ for nonnegative integers $a,b$. If both are positive, reducing each by their minimum increases both $p,q$ and strictly increases the <dimension>. The resulting boundary labels occur: antisymmetrize disjoint upper pairs into lower indices, or lower pairs into upper indices, then take the symmetric traceless component. This proves the maximum formula. For $(k,l)=(5,0)$, the fully symmetric <dimension> is 21 but $(3,1)$ has <dimension> 24. For $k=l$, the fully symmetric traceless <dimension> $(k+1)^3$ is the maximum.
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