$M_{total} = 3 \int^L_0{k V \left( \frac{3}{4} + \frac{x^2}{4L^2}\right) } dx$

$= 3 k V \left[\frac{3 x}{4} + \frac{x^3}{12 L^2} \right]^L_0$

$= 3 k V \left[\frac{3}{4} L + \frac{1}{12} L \right] = \frac{5}{2} k V L $

Equating this to the resistive torque, $\frac{5}{2} k V L = 5T$

$\therefore V_{crit} = 2T/(kL)$.

Since the torque is fixed, you might decrease the minimum wind speed by improving the geometry of the blade and thus increasing $k$, or by increasing the blade length. Increasing the number of blades on the turbine would also decrease $V_{crit}$.

The rotational analogy is

Power = Torque $\times$ Angular Velocity

The power produced by the generator is thus $A \omega_g^2$. The 1:50 ratio of the gearbox tells us that $\omega_g$ can be approximated by $50 B k V$, when $V > V_{crit}$.