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Math Help - Need help solving this math problem

  1. #1
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    Need help solving this math problem

    I need help solving a mathematical problem, my own math skills are not the best. Here's the problem..

    My father built a box of sorts, kind of like a piggy bank for us to put our coins and spare change in. Now the box is quite big and I'm wondering approximately how much it will contain once it's full which i guess will happen in about 50 years.

    Here are the measurements for the box and a coin. I'm using just one coin to simplify it...

    Box; 35,4 inches high (90 cm), lid and bottom (depth); 16x16 inches (40 cm)
    Coin; Rim - 1,88 millimeters, Diameter - 25,00 millimeters

    So let's say the stack up perfectly, how many coins would the box contain once full? Please show me how you solved it. Thanks!
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  2. #2
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    Not sure if this is optimal, but I'll conjecture that the best arrangement is to optimally pack coins at each depth layer (and coins do not cross horizontal layers).

    The number of layers of coins is 90cm/1.88mm.
    The number of coins per layer is around (area of box bottom)/(area of coin) * (packing density) = (40cm)^2 / (pi*(25mm/2)^2) * (density) where density is about pi/(2sqrt(3)).
    See Sphere packing - Wikipedia, the free encyclopedia to understand why this is the packing density.

    Final answer is about (number of layers) * (number of coins per layer) = 477 * 295 = 140715 coins.
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  3. #3
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    Thanks!!!

    So this is not correct then?

    number of coins = volume of container / volume of coin

    = (90 x 40 x 40) / (0.188 x pi x 1.25^2)

    = 156039
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  4. #4
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    Quote Originally Posted by simich View Post
    Thanks!!!

    So this is not correct then?

    number of coins = volume of container / volume of coin

    = (90 x 40 x 40) / (0.188 x pi x 1.25^2)

    = 156039
    This is assuming you can pack all the coins in without any air gaps.
    This does give you a quick upper bound though.
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  5. #5
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    Hello, simich!

    \text{So this is not correct then?}

    \text{(number of coins)} \:=\: \text{(volume of container)} \div \text{(volume of coin)}

    . . = \;\dfrac{90\times 40\times40}{0.188 \times \pi \times 1.25^2}\;=\; 156,\!039

    This would be correct if you melted down the coins
    . . and poured the molten metal into the bank.


    The floor of your bank is 400 mm by 400 mm.
    We can fit 16 coins in a row, and make 16 rows.
    There will be 16^2 = 256 coins in one layer.

    We can have: . \left[\dfrac{900}{1.88}\right] = 478 layers.


    Therefore, the capacity is: . 256 \times 478 \:=\:122,\!368 coins.

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  6. #6
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    Quote Originally Posted by Soroban View Post
    The floor of your bank is 400 mm by 400 mm.
    We can fit 16 coins in a row, and make 16 rows.
    There will be 16^2 = 256 coins in one layer.
    You can do better than 16 rows by offsetting the rows of coins a little bit, so the next row uses some of the gap in the previous row.
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  7. #7
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    Hello, snowtea!

    You can do better than 16 rows by offsetting the rows of coins a little bit,
    so the next row uses some of the gap in the previous row.
    Absolutely right!

    I considered placing the coin in a "hexagonal" array,
    . . with which we can have more coins in a layer..


    . . . . \boxed{\begin{array}{c}<br />
\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\b  igcirc\hdots \\ [-2mm]<br />
\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\bigcirc \hdots \\ [-2mm]<br />
\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\b  igcirc \hdots \\ [-2mm]<br />
\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\bigcirc \hdots \\ [-2mm]<br />
\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\b  igcirc \hdots \\<br />
\vdots \qquad\end{array}}


    Then I calculated the number of coins in a layer ... four times
    . . and came up with four different answers. . *blush*

    Maybe you (or someone) will have better luck.

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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by simich View Post
    I need help solving a mathematical problem, my own math skills are not the best. Here's the problem..

    My father built a box of sorts, kind of like a piggy bank for us to put our coins and spare change in. Now the box is quite big and I'm wondering approximately how much it will contain once it's full which i guess will happen in about 50 years.

    Here are the measurements for the box and a coin. I'm using just one coin to simplify it...

    Box; 35,4 inches high (90 cm), lid and bottom (depth); 16x16 inches (40 cm)
    Coin; Rim - 1,88 millimeters, Diameter - 25,00 millimeters

    So let's say the stack up perfectly, how many coins would the box contain once full? Please show me how you solved it. Thanks!
    The answers you have received so far assume that you will be filling this box in an orderly fashion. My experience is that this is very unlikely. Also unless this is a homework problem you will not be using a single size of coin.

    Under these conditions, random fill with a range of coins, what you have is a problem that is best addressed by experiment.

    Another consideration is whether the box would be able to take the loads imposed by filling it with coins.

    CB
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