Maxwell-Boltzmann distribution most probable speed In my thermodynamics lecture we deduced the Maxwell-Boltzmann distribution,
$$
f_{MB}(\vec{p}) = n\left(\frac{1}{2\pi m k_bT} \right)^\frac{3}{2}\exp \left[-\frac{(\vec{p} -\vec{p}_0)^2}{2 m k_b T}\right].
$$
Where $d^3p\, f_{MB}(\vec{p})$, if I understand correctly, gives the particle density of the particles with a momentum close to $\vec{p}$.
Now, to get the most probable speed in the system $\bar v$ we must, according to my professor, compute the velocity associated with the most probable momentum $\bar p$, which is itself given by:
$$
\frac{\partial}{\partial p} [4 \pi p^2 f_{MB}(p)]|_{\bar p} = 0
$$
Why do we add the factor $4 \pi p^2$? Why is the most probable velocity not simply given by  $\frac{\partial}{\partial p} [f_{MB}(p)]|_{\bar p} = 0 $ (which corresponds to the momentum with the highest associated particle density)?
 A: There's an important abuse of notation happening in your question.  Initially, $f_{MB}$ is the probability density associated to the momentum $\vec p$ - which means that the probability of the momentum being in some small volume $\mathrm d^3p$ centered at $\vec p$ is given by $f_{MB}(\vec p) \mathrm d^3p$.
On the other hand, you might be interested in $\mathcal F_{MB}(p)$, which is the probability density associated to the magnitude of the momentum. This means that the probability that the magnitude of the momentum is in some small interval $\mathrm dp$ centered at $p$ is given by $\mathcal F_{MB}(p) \mathrm dp$. It is this function whose maximum occurs at the most probable value for the magnitude of the momentum.
Your confusion lies in the fact that you use the same notation for both functions.  In fact, the probability that the magnitude of the momentum lies between $p$ and $p+\mathrm dp$ is the same as the probability that the (vector-valued) momentum itself lies in the spherical shell whose inner and outer radii are $p$ and $p+\mathrm dp$.  That is,
$$\mathcal F_{MB}(p) \mathrm dp= \int \mathrm d\Omega \ f_{MB}(p,\theta,\phi) p^2 \mathrm dp$$
Since $f_{MB}$ doesn't depend on $\theta$ or $\phi$, the integral simply yields a factor of $4\pi$ and your result follows from there.
A: The equation you have written is the distribution of the momentum vector. Clearly, the most probable vector is $\vec{p}_0$, and if the system is stationary, i.e., $\vec{p}_0=$, then the most probable momentum is zero: a momentum vector has equal probability to point in any direction, and on average it points to nowhere. The most probable vector is not much of interest, as you can see.
Suppose, however, that we want the most probable magnitude of the vector independently of where it points. A momentum vector of fixed magnitude $p$ draws a sphere of radius $p$ whose surface, $4\pi p^2$, is proportional to the number of particles with that momentum. That's where this factor comes from.
