Math challenge involving camels - anyone know how to solve this...?

Here is a way to get just under 1400 kilograms to Damascus. I am pretty sure that my solution is optimal, but I am not in a position to prove it rigorously right now. (Can anyone do better than 1399.77 kg?)

All decimals are approximate (so, when I say later on that 3 * 200.08 = 600.23, this is because I kept more decimal places than I've actually written out in my answer).

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1. First, have the camel take 1000 kg of grass to a point 1000/19 miles along the path. Then, have the camel deposit as much grass as it can there to still just make it back to Cairo. Repeat this process eight more times, until only 1000 kg of grass remains at the start, and then just take this to the 1000/19 miles waypoint and stop there. At this point, the camel has made a trek of 1000/19 miles nineteen times (nine times there and back, and one time just going forward the last time), so it has eaten 1000 kg of grass. Now 9000 kg of grass remain, at a position 1000/19 miles = 52.63 miles along the path.

2. Next, have the camel take 1000 kg of grass to a point 1000/17 miles further along the path (so it will be a total of 1000/19 + 1000/17 miles away from Cairo). Then, again, it deposits as much grass as it can to still make it back to the 1000/19 miles point, and returns. Again, repeat this until only 1000 kg of grass remains at the 1000/19 miles point, and take all this grass to the 1000/19 + 1000/17 miles point. In this step, the camel has made a trek of 1000/17 miles seventeen times (eight times there and back, and one time just going forward the last time), so it has eaten an additional 1000 kg of grass. Now 8000 kg of grass remain, at a position of 1000/19 + 1000/17 miles = 111.46 miles along the path.

3. Do the same business to travel an additional 1000/15 miles. Now 7000 kg remain at a position of 178.12 miles.

4. Same business, additional 1000/13 miles. 6000 kg remain at position 255.04 miles.

5. Same business, additional 1000/11 miles. 5000 kg remain at position 345.95 miles.

6. Same business, additional 1000/9 miles. 4000 kg remain at position 457.07 miles.

7. Same business, additional 1000/7 miles. 3000 kg remain at position 599.92 miles.

8. Same business, additional 1000/5 miles. 2000 kg remain at position 799.92 miles.

9. Now, something different: Have the camel take 1000 kg all the way to Damascus, and deposit as much as possible there so that it can just return to the 799.92-mile waypoint. Then have it return to the waypoint, and take the remaining 1000 kg to the finish line. In this step, the camel has traveled a total of 3 * 200.08 miles, so it consumed 600.23 kg of grass, which leaves 1399.77 kg to deliver to Damascus.

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Rationale as to why I think this is optimal:

I. Consider the *farthest* point that the camel can possibly get 9000 kg of grass to. To make it to this point, the camel is going to have to make at *least* 9 round trips and one one-way trip, because it can only carry 1000 kg at a time. To get as far as possible with 9000 kg, we should change directions as few times as possible, because traversing back and forth unnecessarily will waste grass. So the best thing to do is to pick a point and just go there and back, making 19 one-way trips. The point furthest along where we can get 9000 kg in this manner is 1000/19 miles.

II. Similar reasoning applies until there are only 2000 kg left.

III. The reason that I consider the farthest point the camel can get and have multiples of 1000 kg of grass is this. We need to make nineteen one-way trips in the first part (to transport 10000 kg anywhere). However, as soon as only 9000 kg remain, we are suddenly relieved of the need for one round-trip; we now only need seventeen one-way trips to transport the remaining 9000 kg. It seems like setting waypoints at the farthest places we can get with multiples of 1000 kg is the best way to take advantage of the fact that the camel now doesn't need to backtrack as much.

IV. As soon as 2000 kg remain, the reasoning from part III no longer applies, because we can actually get 1000 kg *farther* than Damascus. So we need a new strategy instead. Since we need at least three one-way trips to finish no matter what we do, the best thing to do is just get the grass there as directly as possible.

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I am reasonably sure that this is optimal. I am sorry that I am not in a position to offer a fully rigorous proof at this time.

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This problem seems quite complicated and makes my head hurt. It is an odd mix of discrete and continuous reasoning.

the mathemagician is right

max distance will be covered by successively covering distances

in stages that reduce the grass by one load (1000 kg)

considering reduction from 10,000 kg to 1000 kg in stages,

to and fro trips needed will range from 19,17,....to 3, so

distance = 1000(1/19 + 1/17 +....1/3) = 1133.25553 mi

we only need to cover 1000 mi ,

so add back 133.25553*3 kg to 1000 kg to get

1399.76659 kg

if it was only needed to cross the desert

(i.e. ending up using all the grass)

max distance = 1000(1/19 +1/17 +...1/3 +1/1)

= 2133.25553 mi