Let's represent your starting location as $0$ and the destination as $1000$.
Let $f(x)$ be the greatest amount of fuel that can possibly be transported
to or past $x$ miles from the starting point.
For example, if you pick up $1000$ gallons, drive to $1$ (one mile), drop off $998$ gallons, drive back, repeat the trip to $1$ and back,
and on the third trip out you drive to $100$ where you drop
$801$ gallons of fuel, then you will have transported $2995$ gallons
to point $1$: the $1996$ gallons you cached there and the $999$ gallons
that were in the jeep when you passed $1$ on the third trip from $0$.
You should be able to show that for $0 \leq x \leq 200$,
$f(x) = 3000 - 5x$.
The intuitive reason is that you will either have to pass every point
between $0$ and $200$ five times (three times outbound and twice in the
return direction) have to abandon some fuel without using it;
and the latter strategy will deliver less fuel to points beyond where
you abandoned the fuel.
The previous example that transported $2995$ gallons to or past
point $1$ was therefore optimal, or at least was optimal up to $1$.
It follows that only $2000$ gallons can reach $200$ no matter where you
leave your caches along the way.
You should then be able to show that for
$0 \leq y \leq \frac{1000}{3}$,
$f(200 + y) = 2000 - 3y$.
Moreover, you achieve this by delivering exactly $2000$ gallons of fuel
to $200$, including the fuel in the jeep the last time you arrive
at $200$ in the forward direction,
then making sure you have $1000$ gallons in the jeep each time you
drive forward from $200$.
Finally, for $0 \leq z \leq 1000$,
$f\left(200 + \frac{1000}{3} + z\right) = 1000 - z$.
You achieve this by delivering exactly $1000$ gallons of fuel
to $200 + \frac{1000}{3}$ and then fully loading the jeep with any
fuel you have cached at that point and
making just one trip forward.
The answer is $f(1000)$.
Best Answer
Note first that it never makes sense to have an adult return the canoe because that just undoes a trip, so whenever you send anyone across, there has to be a kid available on the far side to return the canoe. Note, too, that the last trip must be made by two kids: whoever brought the canoe back for the last adult has to stay on the starting side while that adult crosses.
So, it takes four trips for each adult that crosses: two to deposit a kid on the far side and bring the canoe back, and another two for the adult to cross and the waiting kid to return the canoe. Once all of the adults are across, it takes one round trip for each kid except the last two to cross. Thus, the minimum number of trips is $4A+2(K-2)+1$.
Finally, note that there must be at least two kids present in order to get everyone across the river.