= Solution
For a fixed $\mathrm{Re}\gg1$, optimize the gain over the initial wave tilt and a later observation time. From the <energy> profile,
$$
\frac{d\log E}{dT}=-\frac{2T}{1+T^2}-\frac{1+T^2}{\mathrm{Re}}.
$$
Its stationary points satisfy
$$
-2\mathrm{Re}\,T=(1+T^2)^2.
$$
There are two negative roots in the high-Reynolds-number regime: an early <energy> minimum and a later maximum. Write them as $T_-$ and $T_+$. Balancing the large-$|T|$ terms for the early root and the small-$|T|$ terms for the late root gives
$$
T_-=-(2\mathrm{Re})^{1/3}\bigl[1+O(\mathrm{Re}^{-2/3})\bigr],\qquad T_+=-\frac1{2\mathrm{Re}}+O(\mathrm{Re}^{-3}).
$$
Thus the optimum disturbance starts as a sufficiently tightly wound leading wave and reaches its peak slightly before $k_x=0$; finite <viscosity> shifts the peak away from the inviscid swing.
Normalize its initial <energy> at $T_-$. The exact maximum ratio is
$$
G_{\max}=\frac{1+T_-^2}{1+T_+^2}\exp\!\left[\frac{g(T_-)-g(T_+)}{\mathrm{Re}}\right].
$$
The prefactor is asymptotic to $(2\mathrm{Re})^{2/3}$, while $g(T_-)/\mathrm{Re}\to-2/3$ and $g(T_+)/\mathrm{Re}\to0$. Therefore <optimal transient amplification of a viscous shearing wave> gives
$$
\boxed{G_{\max}\sim\left(\frac{2\mathrm{Re}}e\right)^{2/3},\qquad \mathrm{Re}=\frac A{\nu k_y^2}.}
$$
The time between the optimum initial tilt and the peak is approximately $(2\mathrm{Re})^{1/3}/(2A)$. Without <viscosity>, ever larger initial tilts would allow unbounded gain; <viscosity> damps them before they can swing, selecting the finite optimum. The coefficient above uses the paper's <Reynolds number> based on $A$, not one based on the full shear rate $2A$.
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