Question

Challenge #1 — The Penny Problem
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.
pennies: v=(3.14)(0.375^2)(0.061)
v=0.026935312
Cylinder: v=(3.14)(3^2)(11.5)
v=324.99
now divide 324.99/0.026935312= 12065 pennies can fit?
Did i do this right?

Answer 1

pennies are round, so they don't pack nicely (in other words, you will get empty space)
so you found a number that is close to , but larger than the actual number that will fit.

Answer 2

?

Answer 3

your math is ok, and if you it were liquids which "fill up" the entire volume, your number would be exact. But you will not fit 12065 pennies into the cylinder, because they will not pack with no spaces. there will be space, and the spaces will use up room

Answer 4

i typed it into the calculator and got 12065.573

Answer 5

if the pennies were squares with sides = 0.75, and you do the same problem, How many would fit ?

Answer 6

okay now im confused.. would you times it by 6 because there are that many sides?

Answer 7

squares have 4 sides. volume will be (0.75)^2 * 0.061

Answer 8

what do i need to do differently to get the right number of pennies?

Answer 9

phi check math chat some kid keeps spamming boob

Answer 10

***
what do i need to do differently to get the right number of pennies?
****
this is a hard problem. We can estimate the number, but getting the exact number would be a lot of work

Answer 11

so should i just not put as many numbers on 0.026935312?

Answer 12

What you did is OK. I am trying to say that even though the math is OK, the answer you got assumes that the pennies can be put in the cylinder and *use up all the space* with not gaps.

Answer 13

yeah and obviously there will be gaps...but what is the problem asking me to find all the gaps or just to assume

Answer 14

The problem is a "challenge problem" (which I assume means it's hard)

For hard problems, you might not get the exact answer, but you can maybe give a range where the true answer lies (for example).
What you did found a number that is *bigger* than the actual number of pennies that will fit.

In other words, if x is the "true number"
x < 12065

Answer 15

so i found to many pennies?

Answer 16

People say you found "an upper bound"
(which is good)
according to wolfram
http://mathworld.wolfram.com/CirclePacking.html
the pennies have a "packing density" of 0.9069
(Your analysis assumed a packing density of 1 (i.e. no gaps))

0.9069 means the gaps use 10% of the area

Answer 17

I would find how many pennies you need to cover the area of the bottom of the big cylinder.

Answer 18

the area of the cylinder is 9 pi
the area of a penny is pi * (0.375)^2
if there were no gaps, we could fit
\[ \frac{9 \pi}{0.14065 \pi} = 64 \]
each penny has thickness 0.061
so we can fit
\[ \frac{11.5}{0.061} = 188.52459 \]
(that last layer is only 1/2 of a penny's height)
but if it fit perfectly we could fit
64*188.52459= 12,065,5...
or 12065, just like you found.

Answer 19

wow okay thanks

Answer 20

now let's tweak that.

Answer 21

the bottom of the cylinder is still 9 pi
but let's assume we can only cover 0.9069 of that with pennies (the rest are gaps)
(remember, the 0.9069 is the packing density , see link)
now how many pennies fit ?
\[ \frac{9 \pi \cdot0.9069 }{0.14065 \pi} = \]

Answer 22

572???

Answer 23

I get 58.04 , or 58 pennies in one layer

Answer 24

about 90% of 64 (64 assumes perfect packing, no gaps)

Answer 25

how did you get 572 ?

Answer 26

But if we believe 58 pennies in one layer and there are 188.5 layers (which we truncate to 188)
we expect to fit 58*188= 10904 pennies

maybe 58 is 1 too big (we would have to look at how the pennies would fill up the round base), and it's only 57 per layer, so 57*188= 10716

Answer 27

you could try to do the experiment with real pennies and see how many can fit into a circle where 8 pennies are exactly the diameter of the circle.
(because 6 inches for this circle divided by 0.75 inches = 8, i.e. 8 of their coins makes up the diameter of the base of the cylinder)