Corey Camel

Every 1000 bananas over the first 1000 will force you to make two trips over any stretch you choose to move them. Thus, for over 2000 you will be making five trips (back and forth twice, then just "forth" the last trip). 1001-2000 will take three, 1000 and under can be carried in one. I have a hard time seeing how splitting up your trips would be advantageous - so long as you have 3000+ you'll still be shelling out 5*distance bananas for it (same goes for the middle and end). This isn't something I can really prove, though I can't see how it could be different either.

Hence, it would make sense to make your first "leg" as long as possible without going so far that you'll be down to 2000. Since you lose 5 per mile, you'd be down to 2000 after 200 miles. Same the next leg, except you lose 3/mi, so you'll hit 1000 after 333 1/3. From there, take it on home. 1000 will be hit at 533 1/3, so you'll have 1000-533 1/3 to go, 1000 bananas -(1000-533 1/3) is 533 1/3. Thus, 533 1/3 banans make it (though I'm not sure what the going rate is on 1/3 banana a camel has been eating off).

Execution would be to simply grab 1000, go 200mi, dump 600 (keeping 200 to get back on) and head back. Repeat, then take the remaining

1000 to 200mi and stay there. You will (corrrectly) have 600+600+800=2000. Again, grab a 1000 and go 333 1/3 mi to 533 1/3. Dump 333 1/3, again keeping 333 1/3 for the return trip (I'm graciously assuming it's ok to split bananas, otherwise you'll had to sacrifice a bit) and go back. Grab the remaining 1000 and go back to 533 1/3, getting there with 666 2/3 for a total of 1000. Grab your 1000 and head the remaining 466 2/3, arriving with 533 1/3.

There may well be ways to improve on this, though I can't think of any. It may just be that I've locked my brain into this concept, but I can't