Considering the general case, I attempt to answer the companion question: if you have $10 N$ gallons of water, how far can you get into the desert?
Obviously, for $N=1$, the answer is 10 miles.
For $N=2$, you carry the water distance $x$, drop off $(10-2x)$ gallons, and go back. Then you pick up the rest, go forward, refill at the cache, and continue on. This is obviously optimal$^1$ if the amount of water left to pick up is exactly the amount you used to get where you are. In other words, $x = 10/3$. At this point, you refill and continue, for a total of $10 (1 + \frac 13)$ miles.
For $N*10$ gallons, you will make $N-1$ round trips and one one-way trip. Thus, the water you use to move $x$ miles is $(2N-1)x$. This should be an even multiple of 10 gallons, or you are wasting water.
For $N=3$, you can pick either $x = \frac 15$ or $x = \frac 25$. In the first case, you cache 20 gallons of water 2 miles in. In the second, you cache 10 gallons of water 4 miles in. In the first case, you can make it $10 (1 + \frac 13 + \frac 15)$ miles, and in the second, you travel $10(1 + \frac 25)$ miles. The first option is clearly slightly better.
In general, for any $N$, you will want to make the next cache at $x = 1/(2N+1)$ miles. That uses up 10 gallons and you then continue with the next smaller $N$, which is more efficient. Making the caches farther in means you are using up your water on less efficient fractions.
Thus, with $10N$ gallons, you can travel as far as
$$10 \sum_{i=0}^{N-1} \frac 1 {2i+1} \text{miles.}$$
The smallest $N$ that makes it to 20 miles is $N=8$.
$$1 + \frac 13 + \frac 15 + \frac 17 + \frac 19 + \frac 1{11} + \frac 1{13} + \frac 1{15} = 2.02$$
Your total distance traveled is $79.8$ miles, and you have $0.2$ gallons left at the end.
This plan extends to any distance.
$^1$If you don't accept the "obviously optimal" comment above:
If there are $10 N + \epsilon$ gallons at a cache point when you are ready to move forward, then you can make $N$ trips from there, and the $\epsilon$ gallons will be wasted. If there are $10 N - \epsilon$ gallons there, then when you go forward, you will be slightly short on your next leg. Making the cache slightly closer would cost you $\epsilon / (2N+3)$ distance on that leg, but gain you $\epsilon / (2N+1)$ on the next leg, for a net win.
Update I didn't count "trips" the same way everyone else does. Naively, this would take 64 trips, but if you optimized, each trip can go one cache further than the previous. I.e.,
- Trip 1, drop $10*{13\over 15}$ gallons at a distance $10*{1\over 15}$ miles.
- Trip 2, refill at the first cache as you pass it, then drop $10*{11\over 13}$ gallons at distance $10*{1\over 13}$ miles past first cache. When you pass the first cache on the way back, grab $10*{1\over 15}$ gallons needed to make it back to camp.
- Trip 3, refill at first two caches as you pass them, drop $10*{9\over 11}$ gallons at $10*{1\over 11}$ miles past second cache. Grab $10*{1\over13}$ and $10*{1\over 15}$ gallons at the second and first caches, respectively.
- Etc.
Thus, you can get all the way across in $15$ trips, assuming the trip-boundary is only when you change direction, not when you refill. If refilling counts as a trip-boundary, then this method will not yield the optimal number of trips.
Update 2
Rereading the question, I see what I should have been answering is "Measured by distance walked, what is the optimal way to cross the desert?" This solution definitely optimizes that, for a total of $79.8$ miles traveled.
