Derivative of R with respect to p, when $c$ is a constant

\begin{align}
\frac{dR}{dp} = cmn\lambda(1-p)^{n-1} + cmnp\lambda (n-1) (1-p)^{n-2}
\end{align}

When $c$ is treated as a constant, the mode is p = 1/n. Hence, when np>1, R decreases as p increases. 

The above inference does not consider that the normalization parameter $c$ also depends on $p$.

2013 Oct 3: $c$ is actually very close to 1 in most simulations.  So, $c$ can be considered as a constant in most cases. Besides, mode p=1/n clearly makes intuitive sense.