Consider the interval 0 < x < w, where w is the width of the lake. Let y(t) be the position at time t of Boater 1, who begins at the left side of the lake (position x = 0); z(t) the position at time t of Boater 2 who begins at right side (position w).
Let c be the speed of Boater 1. During his first trek across the lake, his position is given by the formula
y(t) = ct
This formula is valid for times 0 < t < w/c. During his return trip, his position is given by
y(t) = w - c(t - w/c) = 2w - ct
Similarly, let d be the speed of Boater 2. During his first trek across the lake (from position w to position 0), his position is given by
z(t) = w - dt
This formula is valid for times 0 < t < w/d. On the return trip his position is given by
z(t) = d(t - w/d) = dt - w.
Now let T be the amount of time until the two boaters meet up the 1st time. Then (c + d)T = w, since the combination of distances covered by the two boaters is one width of the lake. Some other constraints we know that hold:
T < min(w/d, w/c),
which means that
w - 600 = y(T) = cT and w - 600 = z(T) = w - dT.
It's not entirely clear to me that we can conclude anything about the time when the two boaters meet up again. It seems possible to me that one boater could be going so slow that the other overtakes him (again) before he has even completed his first trek across the lake. I don't immediately see how to rule out that possibility. Nevertheless, it seems likely to me that the problem's writer is assuming both boaters are on their return trips when they meet again, which would mean that their combined travel distance is 3w at that point, and hence the elapsed time at their 2nd meeting is 3T. That provides us with a couple more constraints:
3T > max(w/d, w/c),
and so
200 = y(3T) = 2w - 3cT and 200 = 3dT - w.
I haven't solved it out, but think it now likely there would be enough constraints (equations) to find the values of all variables c, d, T and w.
