| The Trip around the Moon |
| Analysis: The moon has a circumference of 6790 miles, which means that one unit of fuel has to allow the car to make it 1358 miles (one fifth of the way). We�ll assume that, even though it�s probably NASA heading this expedition, the car is somewhat fuel efficient, allowing for 30 MPG when using regular unleaded gasoline. This means it�ll take 45.27 gallons of to make one unit of fuel, but we�ll round it up to 46. The extra 0.73 gallons will be an allowance for the extra weight from carrying an extra can of fuel to leave in a certain spot.
Now as I write this, the Weigel�s a couple blocks from my apartment has regular unleaded gasoline for $2.399. Using this price and the equivalence of a unit of fuel, it would cost the following amounts for each Method: Method 1 = $8938.68 (actually it�s .674 and not .68, but I�m sure they don�t round down) Method 2 = $17766.99 Method 3 = $19753.37 Even though these prices include tax, I�m sure that NASA would still find a way to skyrocket (teehee) the cost of fuel, alone, to at least $34 million. Will, who posted this puzzle on his blog, said about this puzzle, �This puzzle is in a class called operational puzzles. I doubt you can answer it in closed form with an equation.� Whatever. It took me 3 tries before I found a decent method with which to solve it. Also, my birthday is on the 27th (not this month). 3 * 27 = 81 (The solution of Method 1, and the most fuel efficient way I discovered.) C�mon, dude. Just use your brain. Seriously, though, I noticed that the number of units of fuel it required for Method 1 to place cans at the various locations (2, 6, 14, 34, and 82 for L5) might be points on a fourth-order polynomial (don�t ask me how I noticed, as I�m already coming off nerdy enough as it is). I opened up Excel for the first time since I�d started working on this puzzle and I discovered I was right (R-squared value of 1). I was happy with the fact, even though it meant I�d reached a new level of dorkitude. I tried the next few data points (meaning P6, P7, and P8 while going through L5) and I ended up with an R-squared value of 0.9969 with a fourth-order fit. I�d say not too shabby, though I don�t know if it�s relevant. Anyway, I don�t know that my Method 1 is the most efficient way for the car to get around the moon, but I think it�s a fairly low number, given all that must be done to do this. |