Question:
I have to do a write up that includes a problem statement, process, solution, extensions and evaluation. The problem is: Corey Camel'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? As a reminder we're told to save our notes as we work on the problem. I can't get past an answer of 0. Corey can only carry 1000 bananas and he eats one for every mile so by the time he gets to the market he has 0 bananas to sell and 0 bananas to eat to get him home. Help....I don't know how to do this.

Hi Marie,

I think the answer is 533:

Corey starts with 3000 bananas.

She cannot bring more than 2000 bananas to a distance of 200 miles. Because more would involve three trips forth (and two trips back) which would cost 5 bananas on each mile, for a total of 5 bananas * 200 miles = 1000 bananas, leaving her with no more than 2000 bananas.

But she can bring 2000 bananas to a distance of 200 miles:

1st trip load 1000, eat 200, drop 600, bring back 200 for the road,
2nd trip the same.
3rd trip load 1000, eat 200, drop 800.

Then starting with 2000 bananas, she cannot bring more than 1001 bananas to a distance of 333 miles, because she does one trip forth, one back and one forth, so that each mile costs 3 bananas.

So Corey can get to a distance of 1000 - 200 - 333 = 467 miles of the market with 1001 bananas.

My advice is then to relax and have a banana. Then she can load the 1000 remaining bananas on her back and head to the market, and she'll have 533 bananas when she gets there. (Any other ploy would involve one trip back, hence cost a banana anyway.)

Cheers,
Claude