= Solution
At the optimum, $d=q_*^{1/3}$ and the initial direction is $(1,\rho^+)$. Since the final direction has ratio $\rho^-$,
$$
G_{\max}=q_*^{2/3}\left(\frac{\rho^+}{\rho^-}\right)^2
\frac{1+(\rho^-)^2}{1+(\rho^+)^2}.
$$
Using $\rho^\pm=2R(1\pm s)$, $R^2(1-s^2)=1$, and $q_*=(5-3s)/(5+3s)$ simplifies this to \b[the exact maximum gain]:
$$
\boxed{G_{\max}=\frac{1+s}{1-s}\left(\frac{5-3s}{5+3s}\right)^{5/3},\qquad s=\sqrt{1-R^{-2}},\quad R>1.}
$$
As $R\to\infty$, $s=1-1/(2R^2)+O(R^{-4})$, so $q_*\to1/4$ and $(1+s)/(1-s)\sim4R^2$. Consequently \b[the requested large-Reynolds-number transient growth scaling is]
$$
\boxed{T^*\sim\frac{4\log4}{3}R,\qquad
G_{\max}\sim4^{-2/3}R^2,\qquad
n=1,\ A=\frac{4\log4}{3},\ m=2,\ B=4^{-2/3}.}
$$
These are the <optimal time and gain of a Reynolds-scaled triangular model>. The leading gain can also be read from the off-diagonal <matrix exponential> entry: with $\tau=t/R$, $b=(4R/3)(e^{-\tau/4}-e^{-\tau})$ is maximized at $\tau=4\log4/3$ and has leading maximum $R4^{-1/3}$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-331-optimal-gain.png]
{title=Optimal transient energy gain scaled by Reynolds number squared, approaching its large-Reynolds-number limit}
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