Congratulations to Ahmed Adzemovic, David Land, and the mysteriously named “L S,” our co-winners for this month’s puzzle.
Last month's puzzle asked you to find the optimal balance between driving and charging to maximize the fraction of time spent driving. To solve this, we first need to express the time spent driving and the time spent charging.
It takes $k$ Joules to drive one mile, so if we charge the battery to $Q$ Joules, we can drive the distance $d = Q/k$ miles between charges. Since the car moves at speed $v$, we'll spend $t = d/v = Q/(kv)$ minutes driving for every time we stop to charge.
As for each stop, we have two kinds of time spent:
The pairing time is simple, it's just $T_\text{pair} = 60$ seconds every time.
The charging is the hard part, since the charging rate $r$ depends on how full the battery is. We're told that the charging speed drops linearly as a function of how full the battery is. For example, if the battery is 25% full, the charging rate is 75% of its maximum.
This can be modeled by $r(q) = r_0(1-Q/Q_\text{max})$ where $r_0$ is the initial charging rate, $Q$ is how much energy the battery currently has, and $Q_\text{max}$ is the total capacity of the battery.
With the charging rate in hand, we can figure out how long it takes to reach a given state of charge.
Starting at zero charge, the rate is just $r_0$ and after a short time $\Delta t$, the charge of the battery will be $\Delta q = r_0 \Delta t_1.$ This means that adding the first $\Delta q$ of charge will take time $\Delta t_1 = \Delta q/r_0$.
However, since the charge has now increased, the rate of charging will fall to $r(q) = r_0(1-\Delta q/Q_\text{max})$ which we have to take into account.
To add the next $\Delta q$ worth of charge, we will need time $\Delta t_2 = \Delta q / r(\Delta q).$
The next bit of charge will take $\Delta q/r(2\Delta q)$ and so on and so forth.
The total time to charge up to charge $Q$ is just be the sum of all these: