Srednicki 11.3 part e) Finding the maximum energy for the electron In part e you are asked to find the differential decay rate,
\begin{equation}
     \frac{d\Gamma_{\mu^- \rightarrow e^- \bar{\nu_e} \nu_\mu}}{dE_e} = \frac{mG_F^2}{ 4\pi^3 } \big( mE_e^2- \frac{4}{3}E_e^3 \big)
 \end{equation}
Then you are asked to determine the maximum energy rate for the electron. Now, seemingly the correct result is found by taking the derivative of this differential decay rate and setting that to zero finding,
\begin{equation}
E_{e-max} = \frac{m_\mu}{2}
\end{equation}
For reference, here is a pdf of the problem with solutions. http://hep.ucsb.edu/people/cag/qft/QFT_Notes_11.pdf
However, the advice that was given to me was that the differential decay rate can not be negative and thus the maximum energy occurs at the value when the differential decay rate itself is equal to zero. In that case you get that
\begin{equation}
E_{e-max} = \frac{3m_\mu}{4}
\end{equation}
My main question: It doesn't seem obvious to me why you would be taking the derivative of the differential decay rate to find the maximum electron energy. The differential decay rate should be integrated over from 0 to max electron energy which is only when it is positive, but not when the differential decay rate is at it's maximum, which is what taking the derivative would find. Can someone justify this? Thank you for the help!

 A: This looks like a perfect storm of miscommunication. You appear overwhelmed by  tangential points and to be construing problems that are not there.
First, independently of the rest/bulk of your problem, you must convince yourself from elementary kinematics that, in the rest frame of the μ, and for the masses of both the neutrinos and the electron being so much-much smaller than that of the μ, call it m, the energy (momentum!) of the electron cannot exceed m/2.
Call that momentum p (energy if you wish: you may include the infinitesimal mass of the electron on these scales, and complicate your problem just a little, after you handle the instructive massless electron case). It must be balanced off by the sum of the momenta of the neutrinos, and so the three products lie on a plane. So the projections of the neutrino momenta on the line of this p must add up to p, and the limited energy available for motion is just
$$
m= p + {p\over \cos\theta },
$$
where the second term is the sum of the energies (unprojected momenta!) of the neutrinos. It is then evident that p is maximized by $\theta =0$, so collinear neutrinos, to p=m/2. (It is minimized by the freak situation where the electron is also at rest and the neutrinos are back to back!)
So the normalized electron energy variable is
$$
\epsilon \equiv E_e/(m_\mu/2)= 2E_e/m_\mu,
$$
ranging from 0 to 1, a better variable for your electron spectrum figure, where  that cuts off abruptly.
In these variables, the spectrum is
$$
\frac{d\Gamma}{d\epsilon}\propto \epsilon^2 (3-2\epsilon) ~,
$$
which you have already plotted, with small corrections reflecting details we skipped, angular/helicity twists, etc.
You already see that your $\epsilon =3/2$ is a canard, since it is kinematically impossible to exceed $\epsilon =1$ in the first place. Since the expression arose from a covariant integral with energy-momentum conservation built-in, it is imposible to violate the above kinematic limit, or else a mistake was made in the calculation!
The advice you mention given to you is the observation that the spectral function is monotonic and cuts off abruptly at the kinematic limit $\epsilon =1$, so the maximum is right next to the cutoff zero. The maximum is also the maximum of that function, visible by inspection. You may fuss the details of the cutoff and its location by smoothing the spectral function with realistic corrections, but these are details; basically the maximum is where the function has to smoothly turn over to cut off smoothly, this being physics, virtually at the end of the domain of
$\epsilon =1$.
To give you your due, however, you do have a point in objecting to the strange logic involved. Imagine a freak hypothetical world in which, for the sake of fanciful argument, you got, instead,
$ d\Gamma /d\epsilon \propto \epsilon^2 (1-\epsilon) ,$ with a maximum at $\epsilon =2/3$, well within the kinematic domain, and not at its end. In that case, the problem you are addressing wouldn't really be that meaningful. You certainly have a point in shaking your head at the educational merit of the problem given.
