penny problem: estimate dimensions for holding 1m pennies

[Jolyn Hillen]

Estimate the dimensios of a cubical container needed to hold one million pennies. A penny has the dimensions: diameter .75" and thickness .06"

I have a container that is 33" x 33" x 33" and got it to hold 1,064,800 pennies.

It would have 44 pennies in the rows and columns and 400 high.

Is this correct and is there a way I can get closer to 1,000,000?

 

[galactus]

I don't know how mauch thought is required to go into this. If the pennies are stacked, then there are the gaps between the pennies to consider. Especially since the pennies are round and the container is cubical.

You'd probably be better off just dumping them in.

A penny has volume \(\displaystyle \L\\{\pi}(0.375)^{2}(0.06)=0.0265...\) cubic inches.

If you have a container 30"x30"X30"=27000 cubic inches.

27000/0.0265=1018592 pennies.

This is as if there are no spaces at all. As if all the space is used up. But we always have spaces. I would say a 30" cubic container is pretty close to holding one million pennies.

The wau you done it by stacking, you would have 44^2 pennies to cover the bottom. Since 33/.75=44.

33/.06=550

So, you would have 1,064,800 pennies. As you said.

 

[soroban]

Re: penny problem

Hello, Jolyn Hillen!

Estimate the dimensios of a cubical container needed to hold one million pennies.
A penny has the dimensions: diameter 0.75" and thickness 0.06"

I have a container that is 33" x 33" x 33" and got it to hold 1,064,800 pennies.

It would have 44 pennies in the rows and columns and 400 high.
. . (I think you meant 550 high.)

Is this correct and is there a way I can get closer to 1,000,000?

A cube with side 32½" is closer to one million.

The base would be a square of \(\displaystyle 43\,\times\,43\) pennies
. . and it would be 541 pennies high.

We'd have: \(\displaystyle \:43\,\times\,43\,\times 541 \;=\;1,000,309\) pennies.

 

[Denis]

Re: penny problem

I have a container that is 33" x 33" x 33" and got it to hold 1,064,800 pennies.
Is this correct and is there a way I can get closer to 1,000,000?

Throw away 64,800 pennies :wink:

 

[TchrWill]

Estimate the dimensios of a cubical container needed to hold one million pennies. A penny has the dimensions: diameter .75" and thickness .06"

I have a container that is 33" x 33" x 33" and got it to hold 1,064,800 pennies.

It would have 44 pennies in the rows and columns and 400 high.

Is this correct and is there a way I can get closer to 1,000,000?

Place 4 pennies forming a square with a 5th penny in the center touching each of the outer 4 pennies.

Increase this pattern to the point where 1,000,000 pennies are contained in the cube formed.

Letting n = the number of pennies on each diagonal, the width of the square is W = (n - 1)sqrt(2)/2 + .75

The number of pennies in the square is N = n + 2(n - 2)^2

The total number of pennies is therefore

T = [n + 2(n - 1)^2] x [((n - 1)sqrt(2) + 1.5)/2(.06)= 1,000,000

Solve for n, W and W^3

It may, or may not, be the smallest box.

 

[Deleted member 4993]

Line up 1000 pennies in a row

Then line up 999 pennies in a row above (in same plane) - in hexagonal packing

Then line up 1000 pennies in a row above (in same plane)- in hexagonal packing

Repeat above for 1000 rows

No you have (1000000 - 500) pennies in close pack

Area enclosed calculation

Each row is separated by (0.866 * 0.375 = ) 0.649519053 inches apart

So total height of the rows = 0.649519053 * 999 + 0.75 = 649.6195338 inches

If we stack those extra 500 pennies on the row above , then height =650.2690528 inches

total length of the row = 1000 * 0,75 = 750 inches

toatl volume = 650.2690528 * 750 * 0.06 = 29262.10738 inches

So we need a box of the size 650.3 * 750 * 0.06 = 29263.5 cu in (less than 33*33*33=35937)

 

[TchrWill]

Estimate the dimensios of a cubical container needed to hold one million pennies. A penny has the dimensions: diameter .75" and thickness .06"

I have a container that is 33" x 33" x 33" and got it to hold 1,064,800 pennies.

It would have 44 pennies in the rows and columns and 400 high.

Is this correct and is there a way I can get closer to 1,000,000?

Subhotosh Khan's hexagonal stacking would be the logical packing approach but it cannot result in a perfect "cube" which was the original target.

After some number manipulations, I came up with the following::

1--Start with a horizontal row of 44 coins..
2--Lay a second row of 43 coins atop these 44, each touching the two coins below in the first row.
3--Continue is this manner until you have 51 horizontal rows, 26 44's and 25 43's.
4--These just fit in a 33" by 33.225" area (.75(44) by .6465(50) + .75) making the base size 33" by 34'.
5--The number of coins in this base layer is 26(44) + 25(43) = 2219
6--With 2219 coins to the layer, we need 1,000,000/2219 = 450.65 or 451 layers.
7--The height of the box is therefore 451(.06) = 27.06" or 28" high.
8--This is as close to a perfect cube you can get, I think.
9--The resulting volume is 33x34x28 = 31,4162 cub.in.

The closet other one I found was 30x31x33 for a volume of 30,690 cub.in. One layer is 40 by 47 coins. The base dimensions are 40(.75) by .6495(46) + .75 = 30 by 30.627 or 30 by 31. The number of coins in the base plane is N = 24(40) + 23(39) = 1857. The height becomes 1,000,000(.06)/1857 = =32.31 or 33.