Maximizing power, current, voltage

Well, we know that the power through a resistor is given by:

$$text{P}_text{R}left(tright)=text{V}_text{R}left(tright)cdottext{I}_text{R}left(tright)tag1$$

Ohm's law, states:

$$text{V}_text{R}left(tright)=text{I}_text{R}left(tright)cdottext{R}tag2$$

---

In your circuit, we know that the voltage across the load resistor is given by:

$$text{V}_{text{R}_text{L}}=frac{text{R}_text{L}}{text{R}_text{L}+text{R}_text{s}}cdottext{V}_text{s}tag3$$

The current in the circuit is given by:

$$text{I}_{text{R}_text{L}}=text{I}_text{s}=frac{text{V}_text{s}}{text{R}_text{L}+text{R}_text{s}}tag4$$

So, the power in the resistor $text{R}_text{L}$ is given by:

$$text{P}_{text{R}_text{L}}=frac{text{R}_text{L}}{text{R}_text{L}+text{R}_text{s}}cdottext{V}_text{s}cdotfrac{text{V}_text{s}}{text{R}_text{L}+text{R}_text{s}}=frac{text{R}_text{L}text{V}_text{s}^2}{left(text{R}_text{L}+text{R}_text{s}right)^2}tag5$$

In order to calculate $text{R}_text{L}$ to find the maximum, we find:

$$frac{partialtext{P}_{text{R}_text{L}}}{partialtext{R}_text{L}}=0spaceLongleftrightarrowspacefrac{left(text{R}_text{s}-text{R}_text{L}right)text{V}_text{s}^2}{left(text{R}_text{L}+text{R}_text{s}right)^3}=0spaceLongleftrightarrowspacetext{R}_text{L}=text{R}_text{s}tag6$$

Then we get:

$$text{P}_{text{R}_text{L}}=frac{text{R}_text{s}text{V}_text{s}^2}{left(text{R}_text{s}+text{R}_text{s}right)^2}=frac{text{V}_text{s}^2}{4text{R}_text{s}}tag7$$