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Start by letting L = 4r. You then have a function in two unknowns. The maximum will be at points where the partial derivatives are 0 and the Hessian is negative definite, or if you can show the function is concave at that point.
$\displaystyle \frac{dv}{d\theta}=\omega r \cdot \left(\cos \theta \sin\2 \theta +2 \cos 2\theta \left(\sin \theta +\frac{r}{2L}\right)\right)$
$\displaystyle \frac{dv}{d\theta}=0 \Rightarrow \cos \theta \sin\2 \theta +2 \cos 2\theta \left(\sin \theta +\frac{r}{2L}\right)=0 $
for $\displaystyle L=4r$ you have :
$\displaystyle \cos \theta \sin\2 \theta +2 \cos 2\theta \left(\sin \theta +\frac{1}{8}\right)=0$