Intuition Behind the Maximum Power Transfer Theorem The Maximum Power Transfer Theorem states that

"maximum power transfer occurs when the resistive value of the load is equal in value to that of the voltage sources internal resistance allowing maximum power to be supplied"

Is there a logical explanation for why this is the case? I am aware of the calculus-based proof.
 A: It is possible to show it with algebra only.
Suppose we have a circuit with external and internal resistances equal. In this case $R_e = R_i$, and the dissipated power in the external resistance is:
$$P_e = R_eI^2 = R_e\left(\frac{V}{R_i + R_e}\right)^2 = R_i\left(\frac{V}{2R_i}\right)^2 = \frac{V^2}{4R_i}$$
Now we increase the external resistance:
$$P_e = (R_i + \delta R)I^2 = (R_i + \delta R)\frac{V^2}{(2R_i + \delta R)^2} = (R_i + \delta R)\frac{V^2}{(4R_i^2 + 4R_i\delta R + \delta R^2)}$$
But if we remove the term $\delta R^2$ from the denominator, the fraction is bigger:$$(R_i + \delta R)\frac{V^2}{(4R_i^2 + 4R_i\delta R + \delta R^2)} < (R_i + \delta R)\frac{V^2}{4R_i(R_i + \delta R )} = \frac{V^2}{4R_i}$$
So, we have a smaller dissipated power when increasing $R_e$. The derivation is totally similar to decrease the external resistance, and also shows the same decrease in the power.
The situation with equal resistances is therefore a maximum for the transfer of power.
A: You're apparently already aware of the calculus proof for maximum power $P_L$ delivered from a voltage source, having source resistance $R_S$, to a load resistor $R_L$, is
$$\frac{dP_L}{dR_L}=0$$
giving
$$R_{L}=R_S$$
Where $P_L$ is expressed as $I^{2}R_L$.
As has been pointed out, the calculus is the logical approach to determine the relationship between the load resistance $R_L$ and source resistance $R_S$ to obtain maximum power to the load. But perhaps it would be helpful intuitively to consider what would happen to the power to the load if $R_{L}\ne R_S$. To demonstrate this we can express the power to $R_L$ as the product $VI$ where $V$ is the voltage across $R_L$ and $I$ is the current through $R_L$, and observe the effects on that product when $R_{L}\ne R_S$.
If $R_L$ is less than $R_S$ you will find that the current to $R_L$ increases, but the voltage drop across $R_L$ decreases by a greater amount so that the product $VI$ is less than that when $R_{L}=R_S$.
If $R_L$ is greater than $R_S$ you will find that the voltage drop across $R_L$ increases but the current to $R_L$ decreases by a greater amount so that the product $VI$ is again less than that when $R_{L}=R_S$.
Hope this helps.
A: Here is how engineers visualize this, which is just as Bob D describes it except for a cheat which skips the calculus.
Power dissipated in a load resistor is voltage times current. The power delivered to a resistive load is hence maximized when the product of voltage times current is a maximum. Engineers are taught that this maximum product occurs when the voltage and the current are numerically equal. The proof of this is commonly assigned as an extra credit homework problem in an electrical engineering course, which assignment engineers always skip and instead they just accept this fact as a given because they hate calculus and would rather spend their time tinkering with the carburation on their motorcycles.
For the case where we then fix the source voltage and divide the load resistor into two separate resistors (source and load) in series, we similarly find that the power dissipated in the load is a maximum when the two resistors are numerically equal, R(load) = R(internal).
A: Consider the extremes for a simple Joule-heating scenario, which serves as a fine example of power transfer.
Joule heating at the load is proportional to the current and the load resistance (specifically, the volumetric heating is $J^2\rho$, where $J$ is the current density and $\rho$ is the material resistivity).
Thus, a load resistance down near zero doesn't produce much load heating; there's little for the electrons to bang into to dissipate heat. The limit of zero resistance is zero heating.
Nor does a load resistance up near infinity produce much load heating; the current coming out of the constant-voltage supply is squeezed to nearly nothing. The limit of infinite resistance is again zero heating.
Thus, there must be a happy medium for the load resistance, and there's only one other resistance in the circuit for it to be related to: the voltage-supply internal resistance.
Now, this symmetry argument doesn't preclude the resistances in the optimal state from being related by a factor of 2 or $e$, for instance. The calculus proof is more precise and shows that this factor is exactly 1.
