= Solution
For the linear evolution from part (iii), maximize the energy ratio at each time. The <optimal energy amplification of a linear system> is
$$
G_{\rm opt}(t)=\max_{x_0\ne0}\frac{\|M(t)x_0\|^2}{\|x_0\|^2}
=\sigma_{\max}(M(t))^2
=\frac{S+\sqrt{S^2-4a^2d^2}}2,\qquad S=a^2+b^2+d^2.
$$
Thus the optimizing initial state is a <right singular vector>, not generally an <eigenvector> of $L$.
There is a convenient exact way to find the maximizing time. Let $x_0$ be a unit optimizing initial vector and $y=M(t)x_0$. The <singular value decomposition> gives $M^Ty=G_{\rm opt}x_0$. For $t>0$, the top singular value is simple. At an interior time maximum,
$$
0=\frac12G_{\rm opt}'(t)=y^TLy.
$$
Since $LM=ML$, this also equals $G_{\rm opt}x_0^TLx_0$. Thus both initial and final orientations are neutral energy-growth directions. The positive entries of $M^TM$ let the top <right singular vector> be chosen with positive components. Its orientation decreases through the growth sector, so it must start at $\rho^+$ and end at $\rho^-$.
Put $q=a/d=e^{-3t/(4R)}$ and $K=4R/3$. The solution gives
$$
\rho(t)=\frac{\rho^+}{q+K\rho^+(1-q)}.
$$
Setting this equal to $\rho^-$ yields
$$
q_* =\frac{\rho^+(1-K\rho^-)}{\rho^-(1-K\rho^+)}
=\frac{5-3s}{5+3s},\qquad s=\sqrt{1-R^{-2}}.
$$
Therefore \b[the exact optimal time for $R>1$ is]
$$
\boxed{T^*=\frac{4R}{3}\log\frac{5+3\sqrt{1-R^{-2}}}{5-3\sqrt{1-R^{-2}}}.}
$$
Here and below $\log$ is the natural logarithm. This is the unique positive stationary time: $q$ decreases monotonically from one to zero, and the neutral-direction condition determines one $q_*\in(0,1)$. Also $G_{\rm opt}(0)=1$, $G_{\rm opt}(\infty)=0$, and $R>1$ admits initial energy growth, so this stationary point is the global maximum. For $R\leq1$, the <instantaneous energy-growth criterion for a linear system> gives no amplification, and the maximum is one at time zero.
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