Given a representation of $su(3)$ labelled by $(p, q)$, is there a way to construct its state of greatest weight? My current understanding of the representations of $\mathfrak{su}(3)$ is as follows:
We can construct 3 $\mathfrak{su}(2)$ subalgebras with step operators $I_\pm, U_\pm, V_\pm$. These maybe be expressed in terms of $I_3$ and $Y$, which are proportional to two simultaneously diagonalised Gell-Mann matrices. States are given by points in the $(I_3, Y)$ plane and we move between states using the step operators. $I_+$ for instance is a step $(1, 0)$.
We may label a representation by $(p, q)$ where $p$ is the number of times $I_-$ may be applied to the state of greatest weight before it vanishes, and similarly q is the number of times $U_-$ may be applied to the state of greatest weight before it vanishes. Through commutation relations of the step operators, we find that the states form a hexagon with alternating sides of length $p$ and $q$. Thus we need the $(I_3, Y)$ coordinates of a single state in order to draw the whole lattice.

Additionally, there is an axis of symmetry about the $Y$ axis from which we get the $I_3$ coordinate of the state of greatest weight $M$ as $p/2$. Is it possible to find the $Y$ coordinate of $M$ in a similar manner ?
As an example, the lattice given by the (2, 1) representation, with circles indicating degenerate states:

I would deduce the $I_3$ value for the top-right state to be 1, so that the top row is $(-1, Y_\text{greatest}), \; (0, Y_\text{greatest}),\; (1, Y_\text{greatest})$
 A: I don't have the formal Lie-algebraic statement you are seeking, but the practical seat-of-the-pants answer is evident: The origin of your Y here (in your normalization it is strangeness), that is, the line of zero Y is not the geometric center of your diagram, but the isoquartet line connecting the vertices defining the widest part of your diamond.
For the formal answer, see this.
Here, you are plotting the D(2,1) deciquintet, with one antiquark and two quarks, whose Young tableau looks like a pistol.

(Weight diagram from here. I-spin blue, U-spin green. The inner states are doubled.)
The top line is the isotriplet  $\bar s dd, \bar s du, \bar s uu$.
The second line is the origin of Y, with no strangeness. It consists of an isoquartet spanning the edges of the diamond, mentioned above,
$\bar u dd, (\bar u u- \bar d d)d, (\bar u u- \bar d d) u  , \bar d uu$ with the isodoublet $(\bar u u+ \bar d d)d, (\bar u u+ \bar d d) u$ collocated with the inner states of the above isoquartet. The origin of axes is empty of states.
The third line, of Y=-1, strangeness -1, consists of an isotriplet
$\bar u  ds,   (\bar u u- \bar d d) s, \bar d  us$
with an isosinglet  $(\bar u u+ \bar d d)s$ superposed on its central state. (So the three inner states form a triangle resting on its tip.)
The fourth line, on the southern border, is the obvious isodoublet
$\bar u ss, \bar d ss$.
This comprises the 15 states of D(2,1).   Y and I3 are motivated by strangeness physics, so the diamond is oriented "unnaturally", as you chose, and  its origin for them  is not a barycenter or something! From the convexity of these weight diagrams, it appears sensible to place the origin of Ys in the middle of the isomultiplet with the greatest width: Here, on the 2nd line; or, for the baryon decouplet D(3,0) on the top line, etc...
