You Do The Math - High School Math

I recently saw an anti-education video that stated we don't need no education, specifically, high school math such as the Pythagoraen theorem.  Then I tried to sell some things on eBay.  There you must figure out the weight of your package, the size of the box and the speed you want it shipped.

Doesn't fit.

Easy, right?  Just measure the item to be shipped and find a box that size.   Sure, if you like spending more money on a bigger box.  Here's the list of box dimensions at the post office and the prices for mailing it to a chosen zip code:

The item I needed to ship this time was 14 inches long, and about 3 by 3 inches, width and height.  Clearly, I don't need a box that is 23 by 11, that's really overkill.  But, it's a low price, and for that same low price I can use a 12 by 12 by 5 inch box.  Is that big enough?

Here's a really cool animated gif proving Pythagorus' theorem:

http://en.wikipedia.org/wiki/Pythagoras#/media/File:Pythag_anim.gif

Using the Pythagorean theorem, the diagonal dimension inside the box is equal to the square root of the sum of the squares of the length and width.  The diagonal squared equals 12 squared plus 12 squared.  A gross is 12 squared, or 144.  So, the square root of 288 is the box's diagonal: 17 inches.  Or 16.97 inces if you're being picky.

Doesn't fit.

So, yes, even with padding, my 14 inch item will fit nicely in a 12 by 12 by 5 box for $15.80.
Can I save $4.50 by going with the next size down?   It's 13-5/8" by 11-7/8" by 3-3/8".  If you want almost no padding on the 3" thickness, then let's see if the other dimensins are good enough:

13-5/8" squared  + 11-7/8" squared = diagonal squared
13.625 ^ 2 + 11.875 ^ 2 = 326.66 = 18.07 ^ 2

Yes, there are 4 spare inches along the diagonal.  (Not square, spare.)

Can we use the 11" by 8.5" by 5.5" for the same price, and maybe get a little more room for padding?  No, the diagonal is 13.9 inches.  Very close, though.

Fits with room to spare for padding.

If you notice further down the list, there's a box that the letter carrier will pickup at your house.  Sure, it costs a dollar more than the biggest box we started with, but since you don't have to drive, that saves 5 or 10 dollars for the round trip. (Yes, cars and trucks can easily cost $1 a mile.)

This box is 12" by 10-1/4" by 5", and the diagonal is 15.8 inches.

Either way you go, your high school education saved you about 5 bucks.  Imagine if you shipped ten thousand of these items a year.  That's enough savings to hire a college grad to do all this math, and their salary could pay off their student loans.

How about cubical boxes instead of flat boxes?   What's the smallest cube we could use?  simply square 14 to get 196.  Divide that by two to get 98, and take the square root of that, which is 9.899 inches, just make it an even 10 inches.  This is the measurement of one edge of the box, and the adjacent diagonal on the outside of the box, shown in red in the drawing.

Since the diagonal is 10 inches, we need to use Pythagorus' theorem again to find the remaining edges of the box.  This is the square root of half of 100, which is the square root of 50, which is 7.07 inches.  So, our minimally sized box is 10 by 7 by 7.
 
Just by guessing, it looks like an 8.5 inch cube would work.  And it does, the inside diagonal ends up being just over 14 inches.  The 8.5 inch box looks so much smaller than a 14 by 14 by 14.

Now, if only someone made a 14 by 3 by 3, imaginge how much cardboard and trees could be saved.