N people can walk or drive in a single two-seater car to go from city A to city B. What is the minimum time required to do so?

Assume:
1) the speed of the car: c
2) the speed of a walking person: p
3) the distance between cities: D
4) all persons start out from A at the same time t = 0.
5) all persons arrive in city B at the same time t = T.
6) nobody stands around idly waiting.
7) the car never holds more than two people

Not sure why this is a calculus problem, as algebra is sufficient to solve it.

While there are many ways to accomplish this, assume that one person does all the driving.  He shuttles the other n-1 people between a forward pack that will arrive on foot at the destination at time T and a rear pack that departed the starting point on foot at time 0.

The car drives back and forth a total distance of cT, and advances a net distance of D, so it drives in the forward direction (carrying one passenger) for a total distance of cT/2 + D/2, and it drives back without passengers for a total distance of cT/2 - D/2.  Total time spent driving forward = (cT/2 + D/2)/c = (T +D/c)/2.

Dividing by n-1, we see that each of the n-1 walkers spends time of (T+D/c)/2(n-1) being driven forward for a distance of (cT+D)/2(n-1).
When not being driven, each of the n-1 walkers walks for the remaining time of T - ((T+d/c)/2(n-1)) and a distance of p(T - ((T+d/c)/2(n-1))).

For each of the n-1 walkers,
D = distance driven + distance walked =
(cT+d)/2(n-1) + p(T - ((T+d/c)/2(n-1)))

Solving for T gives

                (p + (2n-3)c)
T = (D/c)* ---------------
                (c + (2n-3)p)
           
This formula only works for N > 1, c >= p, because otherwise we do not have one person driving others.

It checks out for several special cases:

If n = 2, T = D/c  (one car with two passengers does a single one-way trip)

If c = p, T = D/c  (everybody might as well walk)

If p = 0, T = (2n - 3)(D/c) (Nobody walks.  Car makes (2n - 3)/2 round trips from start to finish)

Posted by Steve Herman on 2008-12-06 10:13:25