## Purposeful Practice & Dandy Candies

August 5th, 2014 by Dan Meyer

Practice may not always be fun, but it *can* be purposeful. Some of my favorite tasks lately *contain* purposeful practice.

For instance, Dandy Candies tells students they’re going to package up 24 cubical candy boxes. It asks them, “Which of four packages uses the least amount of packaging? Which uses the least amount of ribbon?”

This is the usual house style. Concrete imagery. No abstraction. Contrasting cases. Predictions. Students make their guesses. Then they get the dimensions from the video. They calculate surface area and ribbon length. (Ribbon length is a little bit more interesting than perimeter but not by a lot.) They validate their predictions with their calculations.

But then we ask them to find out if *another* package dimension will use even *less* material.

So now the students have to think systematically, tabling out their work so they don’t waste effort finding the surface area of a lot of different prisms.

Contrast that against a worksheet like this, which is practice also, though rather less purposeful:

Where else have you seen purposeful practice?

I’d look to:

- Robert Kaplinsky and Nanette Johnson’s Open Middle project.
- Malcolm Swan’s tasks: impossible points & break 25.
- Bryan Meyer’s scientific notation task.

**BTW**. Given any number of cubical candies, what is the best way to minimize packaging? Can you prove it? I can handle real number side lengths but when you restrict the sides to integers, my mind explodes a little.

A workshop participant gave this algorithm. I have no reason to believe it works. I also have no reason to believe it doesn’t work.

- Take the cube root of the volume.
- Floor that to the nearest integer factor.
- Square root the remainder factor.
- Floor that to the nearest integer factor.
- With the remainder factor, you have three factors now.
- The smallest of all three factors is your height.
- The other two are your length and width. Doesn’t matter which.

Note to self: test this against a bunch of cases. Find a counterexample where it falls apart.

**2016 Apr 8**. A similar task from the *Connected Mathematics Project*.

on 05 Aug 2014 at 1:15 pm1 Matt MillarI’ve always found that the use of open ended questions have been a good way of making practice more purposeful (however had not come across open middle before). My students tend to get a lot more practice than they would “answer questions 1-20 on this worksheet” plus you get the pay off of some practice in reasoning and understanding to a degree whereas the worksheet offers only fluency. The beauty I have found is you can also provide that “sting in the tail” rather than “do more of the same”.

I use professor Peter Sullivan’s book on open ended tasks as a starting point

on 05 Aug 2014 at 1:36 pm2 DannyA workshop participant gave this algorithm. Trying with 9 cubes:

Take the cube root of the volume. (2.xx)

Floor that to the nearest integer factor. (2)

Square root the remainder factor. (2.xx)

Floor that to the nearest integer factor. (2)

With the remainder factor, you have three factors now. (2.25)

The smallest of all three factors is your height.

The other two are your length and width. Doesn’t matter which.

So for 9 sides the answer is 2x2x2.5; I hope the cubes are chocolate so you can melt the ninth one and flatten it!

on 05 Aug 2014 at 6:13 pm3 Why I’m not THAT worried about the future of math education… | thegeometryteacher[…] Dan Meyer’s “Dandy Candies” pushes the envelope on video quality, pushes the same content, and includes it in a blog post that discusses a competitor to “I, We, You.” […]

on 06 Aug 2014 at 3:45 am4 Dan Meyer@

Matt, thanks for the link up to Peter Sullivan’s book. The crowd here in Brisbane mentioned him also. Clearly I need to read up.@

Danny, I appreciate your industry here but a quick correction: 2 isn’t an integer factor 9. The cube root of 9 floored to the nearest integer factor is 1.However, that yields 1x1x9, when in reality 3x3x1 has less surface area. So good example, in any case.

2(1*9) + 2(9*1) + 2 (1*1) = 18+18+2 = 38 un^2

2(3*3) + 2(3*1) + 2(1*3) = 18 + 6 + 6 = 30 un^2

on 06 Aug 2014 at 6:57 am5 David PattersonFor packaging that allows no empty space, and for numbers of cubes up to n=107, I propose that you can follow this algorithm:

1. List prime factorization of n in sorted order

2. If list is less than 3 items, add ones. Done.

3. If list is 3 items, done.

4. If list is >3 items, multiply lowest 2 items together and sort again.

5. Goto 3.

I think this works up until you reach n=108, when the algorithm produces 3-4-9, which is less optimal than 3-6-6.

The completion of this algorithm for larger numbers is left as an exercise to the reader. :)

on 06 Aug 2014 at 10:55 am6 JalenWouldn’t you also lump in many of the problem based tasks from Illustrative Math, Mathalicious and Yummymath as examples of problems with purpose?

on 06 Aug 2014 at 11:48 am7 Dan Meyer@

David, can you expand on step four?@

Jalen, this post is about purposeful practice not purposeful problems. The former is tougher to find, IMO, and involves lots of repetition in the context of some overarching purpose.on 06 Aug 2014 at 12:09 pm8 David PattersonLet’s say you have 48 blocks. The prime factorization is 2*2*2*2*3 so the list is 2,2,2,2,3 when sorted. Since there are more than 3 items, multiply the lowest two factors. Now you have 4,2,2,3. Sort them giving 2,2,3,4. You still have more than 3 items, so multiply the lowest 2, which gives you 4,3,4. Now you have 3 items, so you’re done. The packaging should be 3x4x4.

Looking over a list of prime factorizations, it looks like this method would work until you get to 108. 2,2,3,3,3 -> 3,3,3,4 -> 9,3,4 but 3,6,6 is better. (It seems that maybe the next one that doesn’t work is 144.)

on 06 Aug 2014 at 3:34 pm9 Tim Hartman@david patterson

Without doing any math, it looks as though the optimal solution is the three integer factors with the smallest sum.

If so, it seems like the three numbers should be as close as possible to the cube root, hence step one of the first algorithm.

on 07 Aug 2014 at 7:46 am10 Steven@Dan

The remainder factor for step 3 of the original theorem should be 9 shouldn’t it? The square root of 9 is 3 and so you get your efficient packaging.

Works with 108 and 144 as well.

on 07 Aug 2014 at 7:56 am11 Tommy LingbloomThis is very similar to one of my favorite Connected Math investigations. In that investigation, students are simply given 24 cubes and asked to find different ways to package them into a box (rectangular prism). Same premise but stripped back a little bit. I wonder if the video here doesn’t give too much away to students? I like asking the students to come up with the different combinations that will work. When you see a kid with a length of 3 and a width of 3 who is frustrated that they can’t make 24, they start to see the connection between factors, multiples and volume. Having them record the volume (which they quickly realize is always 24) and the surface area leads to the discussion about optimization. They don’t need to start the problem with any formulas, only a conceptual understanding of what volume and surface area are.

on 08 Aug 2014 at 1:55 pm12 Harry O'Malley@Dan

I love the way that this video expresses, both in the playful nature with which the candies dance around as well as in the brilliant song choice, a sophisticated passion for mathematics. It is clear that you care a great deal about your work.

As for purposeful practice, it is leveraged masterfully by the authors of the Philips Exeter Academy materials to create a curricular masterpiece. By embedding practice opportunities into problem solving situations, they have woven together a curriculum focused almost entirely on concept development, without any loss to skill development. Neither a word nor an inch of space is wasted in those texts. Working through them with the right modeling tools on hand is like having the world of mathematics unfold in space like a magical pop-up book. A world of pure imagination.

on 10 Aug 2014 at 11:46 am13 gasstationwithoutpumpsI don’t see the reason to limit oneself to dense packaging—allowing some holes (adding dummy candies) can reduce packaging enormously, and is routinely used in packaging parts.

The algorithm already fails at n=7, which wants 1×2×4, with surface area 2(2+4+8)= 28, while a 2×2×2 box fits 7 (with a space) and area only 6(4) = 24.

If you only need to solve this for small numbers, then a simple computational approach is attractive: compute volume and area for all small boxes with integer side lengths, sort by area (increasing) and subkey volume (decreasing), and remove any from the list whose volume is less than an entry earlier on the list. This provides a list of the biggest volume you can contain for any given surface area (up to the maximum size box of interest).

The more general question is a Diophantine equation problem, which probably means it is difficult to solve analytically.

on 10 Aug 2014 at 5:33 pm14 Sam ShahI felt compelled to comment *just* to say that when I watched the video, I got chills. There is something so perfect about this — at least to me.

on 12 Aug 2014 at 7:50 pm15 ShelleyNate Burchell uses tactile functions in Geometers sketchpad to help students form functions that meet a particular description. He also uses tactile functions to teach his students about parametric curves, and his activity posted on the Sine of the Times blog is a great example of purposeful practice. (Sorry no links, can’t figure out how to hyperlink in the comments here).

on 18 Oct 2014 at 5:27 pm16 More Efficient Packaging | the radical rational...[…] a few Dandy Candies Gift Wrapping to see their thinking in […]

on 04 May 2015 at 12:03 am17 Maya Quinn>Given any number of cubical candies, what is the best way to minimize packaging? Can you prove it?

Have you ever returned to the integral version of this problem? I do not believe it is amenable to elementary means. (Perhaps it can be out-sourced to a combinatorialist, if you happen to know one.)

MQ//Mathwater

on 04 May 2015 at 9:25 am18 Dan MeyerMaya Quinn:Frequently! Here is the solution I find most promising. I haven’t found an example to contradict it and I don’t have a proof that it works.

1. Take your number of candies.

2. Take the cube root of that number.

3. Find the factor of your number of candies that is

closestto that cube root.4. Find the quotient of the number of candies and the factor from [3].

5. Take the square root of that quotient.

6. Find the factor of your number of candies that is

closestto that square root.7. The third factor is trivial.

8. Use the smallest factor as your height.

See any obvious holes I’ve missed?

on 04 May 2015 at 5:49 pm19 Maya Quinn>See any obvious holes I’ve missed?

I am not sure if “holes” is a generic* joke.

Your response indicates that you are thinking about the package as necessarily being a rectangular prism. To ensure that we are referring to the same problem: The version about which I am curious is more general, and allows the cubical candies to be wrapped one side at a time.

To re-paste your original question,

>Given any number of cubical candies, what is the best way to minimize packaging?

The following image shows that the minimal surface area for 9 cubical candies is not achieved by using a rectangular prism:

http://i.stack.imgur.com/iz1Sv.jpg

That image comes from the following link, which also leads to your blog:

http://mathoverflow.net/questions/179371

MQ//Mathwater

*Despite the risk of over-explaining wordplay: This is, indeed, a “generic”/”genus” mathematical pun.