Question

@ganeshie8 @jhonyy9 @dumbcow 1. For the Penny Problem, how much empty space should exist inside the jar after being filled to capacity with pennies? Why doesn't this amount of space actually exist in the jar?

Answer 1

The penny problem:

The first challenge to complete is the Penny Problem. The radio station is giving the winner of this challenge a prize pack that includes tickets to see his or her favorite band in concert. To start off the challenge, the radio station has placed pennies in a cylindrical glass jar. Each penny is 0.75 inches in diameter and 0.061 inches thick. If the cylindrical glass jar containing the pennies has a diameter of 6 inches and a height of 11.5 inches, how many pennies can fit inside the jar? You must show all work to receive credit.

Answer 2

i got approximately 12,081 pennies

Answer 3

u r right

Answer 4

i get 12,065 pennies

http://www.wolframalpha.com/input/?i=9*11.5+%3D+.375%5E2+*+.061+x

Answer 5

@dumbcow sorry i got that too... i need to answer this question tho

For the Penny Problem, how much empty space should exist inside the jar after being filled to capacity with pennies? Why doesn't this amount of space actually exist in the jar?

Answer 6

the empty space is the remainder. To completely fill the jar you need 12,065.6 pennies
but you can't have .6 of a penny so thats the left over empty space
.6 * volume of a penny

Answer 7

ohhhh ok thank you

Answer 8

yw :)

Answer 9

Actually I think the answer will be a bit less, 12032 to be precise.

http://www.wolframalpha.com/input/?i=x+%3D+floor%2811.5%2F0.061%29*%286%5E2%2F0.75%5E2%29

The reason for the floor() is that you can have at most 11.5/0.061 stacks = 188 (round down). And in one circle (or area) there will be \(3^{2}/0.375^{2} = 64\) coins (regarding surface area).

The reason @dumbcow got the additional coins is because the number of stacks is roughly 188.5 .. if you take the half stack and multiply by 64, you'll get the extra coins (which can never be stacked).

\(\bf{\text{Final note:}}\) Well this is as best as you can calculate using high school algebra, the actual number will be much lower due to packing issues.
http://hydra.nat.uni-magdeburg.de/packing/cci/d5.html <-- for time wasting ^_^

Answer 10

haha i think you are over complicating the problem...most high school math problems simplify real world situations

Answer 11

True, the packing part is overcomplicating the problem.

But, the floor part is NOT :).

Basically you can only have an integer number of stacks (or height) which is 11.5/0.061 = 188.5. Since you can't have half a stack (in terms of height), the max number of stacks is 188

And the answer will then be \(188 \times 3^{2}/0.375^{2} = 12032\)

I honestly do believe what I've done is within the limits of high school math (I'm not that old yet :P only been 4 years since HS)