Problem Statement: Corey the Camel has a small banana grove in the desert, her harvest this year was 3,000 bananas. The market where Corey sells her bananas is 1,000 miles away. Corey has to walk to the market to sell her bananas, for each mile Corey walks, she eats one banana. Corey can only carry 1,000 bananas at a time. In this POW the goal is to find the number of bananas that Corey can get to the market. Process: To find the answer to this POW I did the mini POW, as suggested. I used the same process for POW 13, as I did for the mini POW. The process is as follows: 1. Corey starts the trip with 1,000 bananas.
2. She travels 200 miles, she’s left with 800 bananas. She stashes 600 bananas at 200 mile point, keeping 200 the trip back. 3. Corey picks up another 1,000 bananas.
4. She travels 200 miles, she has 800 left. She then picks up 200 from the bananas stashed. She now carries 1000 bananas and has 400 more stashed. 5. She travels an additional 333 1/3 miles, she’s left with 666 2/3 bananas, she stashes 333 1/3 there (533 1/3 mile point), she has 333 1/3 bananas left. 6. She then travels back 333 1/3 miles to 200 mile point. She has no bananas left, so picks up 200 stashed (leaving 200 still at 200 mile point), and travels back to the grove, 200 miles away. 7. She picks up another 1,000 bananas at the grove.
8. She travels to the 200 mile point, leaving her with 800 bananas, she picks up remaining 200 stashed. 9. With 1,000 bananas, she travels 333 1/3 miles to 533 1/3 mile point, she is then left with 666 2/3 bananas. 10. She picks up all 333 1/3 that were stashed there
11. She’s back at 1000 bananas
12. She makes remaining 466 2/3 mile trip, 1000-466 2/3 = 533 1/3 bananas left to sell at the market. Solution:
a. Corey the Camel will have 533 1/3 bananas to sell at the market. b. Yes, I do think that my solution is the best possible, because if you try stashing less bananas at the 200 mile point, or traveling further...