Consider the case of Corey camel - an enterprising owner of a small banana plantation in a remote desert oasis. Corey's harvest, worth its weight in gold, consists of 3000 bananas. The market place where the stash can be cashed in is 1000 miles away. However, Corey must walk to the market, and can only carry up to 1000 bananas at a time. Furthermore, being a camel, Corey eats one banana during each and every mile she walks (so Corey can never walk anywhere without bananas). How many bananas can Corey get to the market? (i am still working on this one myself so i dont have an answer at this stage)
This must be some trick question because it is ridiculously simple: She gets nothing to the market. If Corey Camel must eat a banana each mile on a 1,000 mile trip, she would've eaten 1,000 bananas by the time she made it to the market place, which was everything she was carrying.
consider taking the bananas not the entire way across. for example i loaded up the camel with 1000 bananas, then took her 250 miles, turned and went back to pick up another load of 1000 bananas, etc etc. so at the 250 mile mark i have 1750 bananas. then from there take 1000 bananas to the 500 mile mark (only having to travel 250 miles), go back and bring the remaining 750 bananas to the 500 mile mark. so at the half way mark (500 miles) you have 1000 bananas. then you can just take the 1000 bananas the rest of the 500 miles to the market and cash what you have left (ie 500 bananas). you might be able to take more bananas by increasing or decreasing the interval in which you bring the bananas to.
in white: I believe 1st & 2nd segments of 200 miles (walked 5 times) and 333 1/3 (walked 3 times) are most efficient but don't have a proof 1. start trip with 1000 bananas 2. travel 200 miles, you're left with 800 - stash 600 at 200 mile point, keep 200 for 200 mile trip back. 3. pick up another 1000 4. travel 200 miles, you have 800 left, pick up 200 from stashed, you now carry 1000 and have 400 more stashed. 5. travel an additional 333 1/3 miles, you're left with 666 2/3, stash 333 1/3 there (533 1/3 mile point), you have 333 1/3 left 6. travel back 333 1/3 miles to 200 mile point, you have no bananas left, pick up 200 stashed (leaving 200 still at 200 mile point), go back the other 200 miles. 7. pick up another 1000 8. travel to 200 mile point, leaving 800 bananas, pick up remaining 200 stashed 9. with 1000 bananas travel 333 1/3 miles to 533 1/3 mile point, you're left with 666 2/3 bananas. 10. pick up all 333 1/3 that were stashed there 11. you're back at 1000 bananas 12 make remaining 466 2/3 mile trip, 1000-466 2/3 = 533 1/3 bananas left at end.
My 2 cent.. 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 remaing 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 see how to possibly get a "second start" with 2000 banans further up or a possible more optimal way then to keep the 2000 and 1000 points as far along as possible. You could of course further split the intervals and/or scramble them up, like, say, split the first 200 into two arbitrary ones or drop bananas at will so long as you don't drop more then you need for the trips, such as drop one banana per mile on the way there, then drop all except one and eat the ones you dropped on the way back. However, I can't see how there'd be a benifit, you'd still be at most 200 along at 2000 and so forth, every bit you carry them further while still making five trips (or three on the middle bit) will cost you two banans/mile and every bit shorter won't get them there. [EDIT] Heh, shows what happens when you leave for half an hour and don't hit refresh before posting - someone else says largely the same thing. [EDIT] Noticed I claimed 1000-533 1/3=477 2/3.. Not so, though it has no bearing on the end result since I didn't use it anyhow.
Corey has to take 3000 bananas to a market, which is 1000 miles across the desert, and she can only carry 1000 bananas at a time at max… Each of 1000 bananas will require one trip fro and a trip back to fetch the next batch of bananas, but the last trip will be only forward 5 minimum trips are necessary before she left with 2000 bananas. 1000/5 = 200 miles covered. 800 miles to b covered with 2000 bananas. 3 min trips are necessary for next 1000 bananas (2 forward and one back) 1000/3 = 333.33… miles covered. Total distance covered at this point = 533.33… 466.66… miles to be covered with 1000 bananas 534 bananas remain at the end. Now reaching destination with 534 bananas has many back trips… but an optimal solution will be the one with minimum back trips, i.e., back trips should tend to zero. For this the distance covered should be calculated in steps instead of miles as considered before. In this scenario each and every step taken will be counted. Corey’s appraisal Target -> Take 3000 bananas to a market, which is 1000 miles carrying 1000 bananas at a time at max… If Corey reaches the destination with less than 534 -> she will be rated as Below Expectation. If Corey reaches the destination with exactly 534 -> she will be rated as Meets Expectations. If Corey reaches the destination with more than 534 bananas -> she will be rated as Above Expectations. * * * <534 =534 >534 (Below Expectation) (Meets Expectation) (Above Expectation) Now the maximum that can be taken to the final point can be calculated as below. The optimal solution is the one where in there are no trips backwards. 3 trips for the first 3000 bananas before the count of bananas is 2000. 333.33… miles covered with 2000 bananas left. Next 500 miles can be covered with 1000 bananas left. 833.33… miles covered with 1000 bananas left. 833.33… bananas left after covering the left 166.66… miles. To be left with 833 bananas at the end of the trip Corey has to travel only in the forward direction carrying all the bananas with her like dragging the bananas along the way. In this scenario each and every step counts. In terms of probability, reaching the final point with 534 bananas has a probability of 1 and reaching with 834 bananas has a probability of 0. And one who accomplishes a task, which has a probability of just greater than zero is always regarded as the best. Now it is left to Corey what she wants to accomplish...
I recognize this problem, it was assigned as a "problem of the week" in my high school algebra class. To find the best answer you must first solve a sub-question which is simpler, then apply the answer to the original question, using proportionalities. The sub-question is: Mini-Camel owns a banana grove like Corey Camel. However mini-camel's harvest is only 45 bananas and he can carry at most 15 bananas at a time. Of course, the marketplace is 15 miles away and, like Corey Mini-Camel must eat one banana each and every mile he walks. How many bananas can he get to market? How does he do this? Hint: Both Corey and Mini-Camel can make "pit-stops" along the way and stash bananas there, then go back to their groves or the last pit-stop. Warning: the answer is NOT 0 for either Corey or Mini-Camel!
Wait a second, if the market is 1000 miles there and a 1000 miles back how does Corey get back to the plantation without any bananas?
Dear "The Riddler," when I first got this probem as a POW in my old high school math class, I thought the exact same thing. So, I walked up to my algebra teacher and asked her. And she told me I was overthinking it. Corey dous not have to make it home...infact, the problem is impossible if he had to! If you or anyone else wants to say, "What!" that's complete BS and unrealistic...then just do what I did, say he came home in a helicopetor! See, he dosn't own a phone at his house, so he can't hire a helicopter but, when he gets to town then he can to get home! Make sure you do the mini-problem I posted, solving it makes the main problem so much easier! spoiler: hello my name is spoiler. Spoiler, Sp-oil-er, spoiler fun! I love spoiler time, let's all get on the spoiler train. This is to give you time to stop reading if you wan't to fid the answer fairand square. Warning you will soon reach the spoiler stop reading write know. here is the spoiler. it's a proportionality, you'll know what that means when you get to the write point. that was a spopiler. hope you had fun in spoiler town. enjoy figuring out the problem but, remember, you can't gloat because you used a spoiler! spoiler, spoiler, spoiler. la la la spoiler. Haa!
500 bananas at the market. Start off with 3000 bananas at home. Markers H 2 5 7 M represent Home, 250, 500, 750 miles, and market respectively. Take 1000 to marker 2. Swallow 250 on the way over and on the way back. Left with 500 @ 2. Take the next 1000 to marker 2. Same amount swallowed. Left with 1000 @ 2. Take the last 1000 to marker 2. Swallow only 250. Left with 1750 @ 2. Take 1000 to marker 5. Swallow 250 on the way over and back. Left with 500 @ 5. Take 750 to marker 5. Swallow 250 only. Left with 1000 @ 5. Take 1000 to marker 7. Swallow 250. Left with 750. Take 750 to marker M. Swallow 250. Left with 500. Actually more economically-sound: Eat about 250 bananas while working out plans to make an aircraft. Fly to market, screw walking, arrive at market with 2750 bananas. Actually I'm trying to figure out if theres a nice formula for this. Then we can just maximize it and get our answer.
nice puzzle i did it on a per mile basis with 3 different stages to the trip GENERAL STRATEGY: corey loads up with the max amount of bananas, (1000) travels 1 miles, (999) dumps all but one banana (998) uses that banana to return to the stockpile and repeats until the stock pile is empty STAGE 1 corey has more than 2000 bananas each mile costs 5 bananas 3 forward trips and 2 reverse trips so corey reaches the 200 mile mark with 2000 banans left Stage 2 corey has between 1000 and 2000 bananas each mile now costs 3 bananas 2 forward trips and 1 reverse trip so corey traves a furter 333+1/3 miles (total 533+1/3) with 1000 banans left stage 3 corey has 1000 or less bananas each mile now costs only 1 banana, 1 forward trip only corey travels the last 467+2/3 miles costing 467+2/3 bananas corey has traveled 1000 miles with 533+1/3 bananas left
Corey walks one mile with 1000 bananas, eats one banana. Corey walks back to plantation, eats another banana. Corey walks back to stash, eats another banana. Corey walks back to plantation and eats another banana. Corey walks back to stash and eats another banana. Looks to me like Corey just ate 5 bananas to move 2995 bananas one mile. It appears this problem is incorrectly posed. If you integrate this as a per 'banana mile' of load, you do get an optimal delivery rate close to 500 bananas.
So if there is no restriction on how much the camel can eat (its "internal" fuel tank), why couldn't the camel eat 1000 bananas all at once, and trot off across the 1000 miles while carrying the 1000 bananas on her back? The bananas would digest (1 banana per mile). That way, the camel would arrive at market with an empty tummy and 1000 bananas for sale?
"Corey eats one banana during each and every mile she walks" It's not quite this explicit, but it effectively specifies that there's no 'stocking up' in terms of eating then walking. If you'd like, you can imagine the problem as "Corey's mouth/esophagus/stomach/intestines cannot hold more than 1 banana at a time, and walking dissolves any whole or partial bananas within Corey at a rate of 1 per mile". More exactly, the OP is saying that somewhere in the process of walking a distinct mile, Corey eats one whole banana. It's not clear if the banana is eaten partially while walking, or at the beginning or end of the mile-- just that if Corey walks a mile, she is guaranteed to have actively eaten a whole banana sometime in the interim. DaveE