jmc

Since it doesn't take too long, we can just make a table with all the possible values of the sum: $$\begin{array}{ccc}\text{multicolumn}3& & \\ ♢& \Delta & ♢+\Delta \\ 36& 1& 37\\ 18& 2& 20\\ 12& 3& 15\\ 9& 4& 13\\ 6& 6& 12\end{array}$$ Clearly $37\to \boxed{37}$ is the largest.