How many glasses with a base of 68 mm will fit the tray of 28 x 42 cm

[Jignesh77]

This glass has vertical sides. The diameter of its base is 68 mm. How many of these glasses will fit of this rectangular tray (28 cm × 42 cm)?

Thank you in advance for your help.
The correct answer is 24 glasses.
Why can't we do the following?

Area of the tray = 28 × 42 =1176 cm2.

Area of the base= pi x r x r = 36.29 cm2

1176/36.29= 32.40.

I saw in my son's year 6 practice book. I know the correct answer is 24 but I still don't know why not 32.40.

 

[Dr.Peterson]

This glass has vertical sides. The diameter of its base is 68 mm. How many of these glasses will fit of this rectangular tray (28 cm × 42 cm)?

Thank you in advance for your help.
The correct answer is 24 glasses.
Why can't we do the following?

Area of the tray = 28 × 42 =1176 cm2.

Area of the base= pi x r x r = 36.29 cm2

1176/36.29= 32.40.

I saw in my son's year 6 practice book. I know the correct answer is 24 but I still don't know why not 32.40.

Can you fit circles together to fill a rectangular space without any extra space between them? That's what you're assuming.

This is not an area question, because you can't reshape the circles to make them fit. The best you can do is to imagine each circle enclosed in a square, and fit those together (unless it happened to be possible to make a triangular array, like a honeycomb -- for these dimension, you could only fit 22 that way).

 

[The Highlander]

This glass has vertical sides. The diameter of its base is 68 mm. How many of these glasses will fit of this rectangular tray (28 cm × 42 cm)?

Thank you in advance for your help.
The correct answer is 24 glasses.
Why can't we do the following?

Area of the tray = 28 × 42 =1176 cm2.

Area of the base= pi x r x r = 36.29 cm2

1176/36.29= 32.40.

I saw in my son's year 6 practice book. I know the correct answer is 24 but I still don't know why not 32.40.

Think about it.

Even if you squeezed all the glasses together (right up next to each other), each glass would occupy a square of 6.8 cm sides because of the 'wasted ' space between them!

That means you could only fit in four down and six across (ie: 24); see pic below.

Hope that helps.

​

 

[mario99]

This glass has vertical sides. The diameter of its base is 68 mm. How many of these glasses will fit of this rectangular tray (28 cm × 42 cm)?

Thank you in advance for your help.
The correct answer is 24 glasses.
Why can't we do the following?

Area of the tray = 28 × 42 =1176 cm2.

Area of the base= pi x r x r = 36.29 cm2

1176/36.29= 32.40.

I saw in my son's year 6 practice book. I know the correct answer is 24 but I still don't know why not 32.40.

If we assume that any extra space inside the rectangular tray can be ignored, then [imath]24[/imath] glasses will fit. How?

First convert the units to [imath]\text{mm}[/imath], so that the dimensions of the rectangular tray is [imath]280 \ \text{mm} \times 420 \ \text{mm}[/imath].

How many glasses are there along the length of the rectangular tray?
[imath]\displaystyle\frac{420}{68} \approx 6[/imath]

How many glasses are there along the width of the rectangular tray?
[imath]\displaystyle \frac{280}{68} \approx 4[/imath]

So there are [imath]6 \times 4 = 24[/imath] glasses.

 

[khansaheb]

Just an extension of the problem:

If the tray size was (68*6=) 408 mm x 612 mm and the glass dia. was 68 mm as before -
then you could fit-in 55 glasses on the tray by using hexagonal packing scheme.
The packing scheme described in response #3 is known as (generally) rectangular packing. Using that scheme, you would fit in 54 glasses (9*6) in the tray.

 

[Dr.Peterson]

Actually, I just did the calculations for the original problem if we use hexagonal packing with the first row spaced out across the 28 cm side (so there is 0.2666... cm between glasses), rather than using regular hexagons with the first row touching, we can fit 25 glasses in the 42 cm length. (The actual distance used is about 41.66 cm.)

So the problem almost appears to be designed for this surprise answer, rather than the obvious rectangular packing: the 42 cm length could have been reduced to 41 cm and still give the expected answer of 24 glasses by rectangular packing, and that would prevent this alternative answer.

Of course, this answer would not be expected at the stated year 6 level.

 

[Dr.Peterson]

Here's what it looks like:

 

[The Highlander]

Here's what it looks like:

Drat! Another hour wasted drawing the situation described (at Post #6).
My picture is prettier though.


@Jignesh77: You should perhaps contact the publisher/author(s) of your son's book (providing a link to this thread) and point out that their answer is wrong; you might also suggest they alter the width (to 41 cm) so that it's no longer possible to fit 25 glasses on the tray?

 

[Dr.Peterson]

My picture is prettier though.

But mine (which also took far too long) is an exact GeoGebra construction (to make sure I was right).

It's conceivable that the intent is, after solving mathematically (but naively), to do a physical demonstration and show that it's possible to fit in an extra one just by jiggling everything carefully.

Circle packing is, in general, a hard problem, not one you can solve routinely, even when an answer seems obvious..

 

[mario99]

Actually, I just did the calculations for the original problem if we use hexagonal packing with the first row spaced out across the 28 cm side (so there is 0.2666... cm between glasses), rather than using regular hexagons with the first row touching, we can fit 25 glasses in the 42 cm length. (The actual distance used is about 41.66 cm.)

So the problem almost appears to be designed for this surprise answer, rather than the obvious rectangular packing: the 42 cm length could have been reduced to 41 cm and still give the expected answer of 24 glasses by rectangular packing, and that would prevent this alternative answer.

Of course, this answer would not be expected at the stated year 6 level.

I have a small question to the genius Dr.

With the arrangement of post #3, done by Highlander, the total ignored extra space occupied by the glass in each row and column = [imath]\displaystyle 4*\frac{2}{17} + 6*\frac{3}{17} = \frac{26}{17} \approx 1.529[/imath]

Does this wasted space mean that there must be a peculiar arrangement that will let 25 glasses fit inside the rectangular tray as you did in post #7? Or it is irrelevant?

 

[Dr.Peterson]

With the arrangement of post #3, done by Highlander, the total ignored extra space occupied by the glass in each row and column = [imath]\displaystyle 4*\frac{2}{17} + 6*\frac{3}{17} = \frac{26}{17} \approx 1.529[/imath]

Does this wasted space mean that there must be a peculiar arrangement that will let 25 glasses fit inside the rectangular tray as you did in post #7? Or it is irrelevant?

Where did that calculation come from, and what does it mean? You appear to be talking about space between the squares around each glass, which is not "extra space occupied by the glass".

And whether there is a way to fit an extra glass doesn't follow from the mere fact that the squares don't fill the space.

As I said, these problems can get complicated. Did you look at the page I linked to which shows best-case fitting like this?

​

 

[mario99]

Where did that calculation come from, and what does it mean? You appear to be talking about space between the squares around each glass, which is not "extra space occupied by the glass".

And whether there is a way to fit an extra glass doesn't follow from the mere fact that the squares don't fill the space.

As I said, these problems can get complicated. Did you look at the page I linked to which shows best-case fitting like this?

View attachment 37712​

I am sorry if I was ambiguous to describe my thoughts. I meant that each ignored space (red and green) occupied by the glass will have a segment left from the diameter equal to [imath]\frac{2}{17}\text{D} [/imath] (green) and [imath]\frac{3}{17}\text{D} [/imath] (red). If we sum these segments, they are equal to [imath]\frac{26}{17}\text{D} \approx 1.529 \text{D} [/imath], which is slightly greater than a glass and a half. ([imath]\text{D}[/imath] is the diameter.)

Where did that calculation come from

These calculations came from:

[imath]\displaystyle \frac{420}{68} = \frac{105}{17} = 6 + \frac{3}{17}[/imath]

And

[imath]\displaystyle \frac{280}{68} = \frac{70}{17} = 4 + \frac{2}{17}[/imath]

Did you look at the page I linked to which shows best-case fitting like this?

View attachment 37712​

Yes, I did scan the link. I don't fully understand it, but with carefully reading it multiple times, I think that I will get the idea.

And whether there is a way to fit an extra glass doesn't follow from the mere fact that the squares don't fill the space.

This point is very interesting and I think that it is the key to say: there is no systematic way to fit an extra glass (circle) for all cases, but we can at least guarantee some of the first few square numbers.

Back to my core question.
I am just wondering if this wasted space (red and green) which is slightly greater than a glass and a half ([imath]1.529\text{D}[/imath]) has anything to do with the [imath]25^{th}[/imath] glass!

Correction!
For the sake of completeness, I should have written:

[imath]4*\frac{3}{17}\text{D} + 6*\frac{2}{17}\text{D} = \frac{24}{17}\text{D} \approx 1.412\text{D} [/imath], which is still more than a glass, and that what matters.

 

[Dr.Peterson]

I am just wondering if this wasted space (red and green) which is slightly greater than a glass and a half (1.529D1.529\text{D}1.529D) has anything to do with the 25th25^{th}25th glass!

Since your "wasted space" relates specifically to space around the squares in the original solution, and any denser arrangement will not use squares, I don't think there's any direct relationship. The important thing for my particular improvement is that there is enough room in one direction to spread out the circles a little, and that that in turn allows things to fit differently.

 

[The Highlander]

But mine (which also took far too long) is an exact GeoGebra construction (to make sure I was right).

Of course your figure is a much more accurate depiction of the situation; I realized that as soon as I saw it.

Unfortunately, I began my own little artwork just after I read your Post #6 and wasn't aware that you had posted your own graphical representation of it until I returned (after dealing with cooking & laundry matters as well as the artistic pursuit) to post my own efforts.

I trust you can imagine my dismay, therefore, when I discovered that you had already posted a (more geometrically correct) diagram; my comment that mine was "prettier" wasn't in any way intended to criticize your graphic, it was just a lighthearted attempt to offer some justification for posting my own.

I'm afraid that (much as I would really like to) I simply don't possess your GeoGebraic skills. I probably could have tried to make a more accurate diagram in Desmos (see pic below) but, as you can see, that would have necessitated constructing a separate equation for every single one of the 25 circles which I found somewhat unappealing. Nevertheless, before starting this post (and as an experimental confirmation of my thoughts on doing it in Desmos), I started it (below) but just couldn't be bothered pursuing it to completion, however, anyone who wishes to finish it is welcome to do so and can access what I started here.

Did your implementation in GeoGebra involve similarly laborious computations (is that why it took longer than you were happy with?) or is it simpler to create the diagram in that environment?

In defence of my own submission, I knew, of course, that it wasn't going to be geometrically accurate (which is why I also included the calculation of the distance between column centres) and, given all the (pertinent) dimensions that I included, I believe that, despite its geometric shortcomings, it would pass muster as a perfectly acceptable sketch of the situation.

It's conceivable that the intent is, after solving mathematically (but naively), to do a physical demonstration and show that it's possible to fit in an extra one just by jiggling everything carefully.

You may be right, ofc, but I suspect you are giving far too much credit to the author(s) of the question.

Drawing it in Desmos...

​

 

[Dr.Peterson]

I trust you can imagine my dismay, therefore, when I discovered that you had already posted a (more geometrically correct) diagram; my comment that mine was "prettier" wasn't in any way intended to criticize your graphic, it was just a lighthearted attempt to offer some justification for posting my own.

I intended to be light-hearted too, though it didn't come across that way without emojis

I'm afraid that (much as I would really like to) I simply don't possess your GeoGebraic skills. I probably could have tried to make a more accurate diagram in Desmos (see pic below) but, as you can see, that would have necessitated constructing a separate equation for every single one of the 25 circles which I found somewhat unappealing. Nevertheless, before starting this post (and as an experimental confirmation of my thoughts on doing it in Desmos), I started it (below) but just couldn't be bothered pursuing it to completion, however, anyone who wishes to finish it is welcome to do so and can access what I started here.

Did your implementation in GeoGebra involve similarly laborious computations (is that why it took longer than you were happy with?) or is it simpler to create the diagram in that environment?

I did it as a construction, placing centers in the leftmost column by specifying coordinates, then placing the next column by 6.8 cm circles, and the rest by intersecting lines. I didn't do any computation beyond adding radii and spaces. (I did it this way partly as a check on my calculations, so I didn't want to use them to do it.) At one point some circles were overlapping, which turned out to be due to a wrong number in the first column, and was easily corrected.

In defence of my own submission, I knew, of course, that it wasn't going to be geometrically accurate (which is why I also included the calculation of the distance between column centres) and, given all the (pertinent) dimensions that I included, I believe that, despite its geometric shortcomings, it would pass muster as a perfectly acceptable sketch of the situation.

It doesn't need defense. I originally wanted to make a sketch to show the idea, but didn't have the patience to do that, or to show the calculations. Together, we made things clear.

You may be right, ofc, but I suspect you are giving far too much credit to the author(s) of the question.

Probably. That's why I chose the word "conceivably". I could imagine it; could they?

 

[The Highlander]

I had a spare half hour today (twiddling my thumbs, lol) and it occurred to me that I had already calculated all the necessary y-coordinates of the circles' centres and only needed a couple more x-coordinates to complete my drawing in Desmos.

So I thought I would just finish it off after all; result below.

(If anyone is interested in seeing how it is constructed in Desmos then it's creation may be viewed here.)

​