Maximal dimension in mixed SU(3) tensor powers ($D_{\max }(k,l)$)

For $d(p,q)=(p+1)(q+1)(p+q+2)/2$, the exact largest constituent dimension of a mixed SU(3) tensor representation isEvery 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.