Why exactly does maximum power transfer happen at 50 ohms (matched impedance)?
The power delivered to the load is from the Joule heating effect: 

\begin{equation}
P=\dfrac{\Big(\dfrac{R_L}{R_L + 50}G\,V_{IN}\Big)^{2}}{R_L}
\end{equation}

So from differential calculus we know that a function reaches its maximum or minimum value when we differentiate and equal it to zero. In this case we'll have a maximum value, thus:

\begin{equation}
\dfrac{dP}{dR_L} = (G\,V_{IN})^2 \dfrac{((R_L+50)^2 - 2\cdot R_L\cdot(R_L + 50))}{(R_L+50)^4}
\end{equation}

Finally by making \$ \dfrac{dP}{dR_L} = 0 \$ we only need to care about the numerator since the denominator does not make the equation be equal to zero from any real value. Thus we have:

\begin{equation}
(R_L+50)^2 - 2\cdot R_L\cdot(R_L + 50) = 0 \implies \, (R_L)^{2} = 2500 \implies R_L = 50\,\Omega
\end{equation}