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As the title mentioned - I am not sure why exactly the maximum power delivered to the load will be max when R_L is 50 ohms.
If I guess why, it's because if the resistance was greater than 50 ohm then the current will be lower, but if it was less than 50 ohm (e.g. 25 ohm), then the constant 50 ohm resistor would deliver the majority of the power instead of going to the load.
Why does the maximum power transfer happen at 50 ohms?
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}
Intuitively: when you raise the load resistance, you are increasing its share of the voltage (and thus power) versus the other resistance; but you are decreasing the total current (and thus power). When you lower the load resistance, you are decreasing its share of the voltage (and thus power); but you are increasing the total current (and thus power).
So which direction power goes, up or down, depends on which effect is stronger. And as it happens, they cross over at 50 ohms (that is, when load resistance is equal to source resistance.)
As a rule, maximum power transfer for the load with a series and load resistor always happens when the resistances are equal. Use this rule as a shortcut when designing anything from antennas or transmission lines.
Another way to solve this, that doesn't involve actually doing any differentation.
Let \$R_S\$ be the source resistance, \$R_L\$ be the load resistance, \$V_{RL}\$ be the voltage accross the load, \$V_{RS}\$, be the voltage across the source resistance, \$P_{RL}\$ be the power delivered to the load resistance and \$I\$ be the current. \$G\,V_{IN}\$ and \$R_S\$ are outside of our control. We control \$R_L\$ and out goal is to maximise \$P_{RL}\$.
$$G\,V_{IN} = V_{RS} + V_{RL}$$
$$P_{RL} = IV_{RL} = \frac{V_{RS}}{R_S}{V_{RL}}= \frac{V_{RS}V_{RL}}{R_S}$$
So to maximise the power to the load we need to maximise \$V_{RS}V_{RL}\$, since this is a quadratic it has exactly one turning point, and by symmetry that turning point must be at \$V_{RS} = V_{RL}\$, which in turn implies \$R_S=R_L\$.
