\( R = p N \int_p^{\infty} f(p) \)

Or to maximize

\( log(R) = log(p) + log( N ) + log[1-F(p)] \)

Now, I take the first order derivative with respect to p, to find the optimal value of p.

\( \frac{\partial log(R)}{\partial p} = \frac{1}{p} -\frac{f(p)}{1-F(p)} =0 \)

This implies that the optimal value of p satisfies the equation

\( 1 - F(p) = pf(p)\)

Now, under the assumption that \( f(p)\) is distributed as uniform or exponential, the optimal value of \( p\) is the mean of the distribution. This also makes sense from intuitively. However, how do I show this for any general (perhaps non-negative) f(p)?

Additionally, integrating both sides of the equation, we get \( E(p) \) on both sides of the equation. I dont think that is very helpful but thought I would point that out.