Rectangular Milk Bottle Crate

The Problem

Qi-xiao has a new problem. He works with a company that uses square-bottomed milk crates but his boss thinks that they might get more bottles into rectangular crates instead. The idea is to divide up the crates with dividers. Using 3 dividers Qi-xiao can fit 4 or 6 bottles into a crate, depending how he uses them. (Both of these situations are shown in the picture.)

On Monday, Qi-xiao’s boss wants to know the largest number of bottles that can be fitted into a rectangular crate if 11 dividers are used.

On Wednesday, Qi-xiao’s boss wants to know the smallest number of dividers that would be needed for a rectangular crate that could hold 80 milk bottles.

Teaching sequence

1. Pose the first part of the problem for the students to think about. As a class share ideas on:
   How could we set this up?
   What information do we know?
   What mathematical knowledge could we apply to this situation?
   How will we compare the cases?
   What is the fewest number of bottles that could be fitted into a crate with 11 dividers? Explain.
2. Pose the rest of the problem. Ask the students if they require any clarification of the problem.
3. While the students are working on the problem ask questions that enable them to clarify the variables involved in the problem.
   What starting strategy did you use?
   What changes in this problem?
   What variables do you need to consider?
4. Encourage the students to write down any connections they have found in words and link this to a possible algebraic form.
5. Share answers and reasoning for each part separately. Consider the range of approaches used.
   Does your answer make sense? Explain why.

Extension to the problem

On Tuesday, Qi-xiao’s boss wants to know the largest number of bottles that can be fitted into a rectangular crate if 2d + 1 or 2d dividers are used. In each case, d is a fixed number.

On Thursday, Qi-xiao’s boss wants to know how many as well as the best-shaped crate that could hold 400 milk bottles. Generalise.

Solution

Possible solutions include using a table and using algebra. We’ll only use a table here but go to the algebraic method to look at the Extension problem.

There are many ways that we can use the dividers. We show these in a table. The ‘dividers’ column show the various sums that add to 11. The length and width of each possible crate are each then one more than the addends in the dividers column. The number of bottles is then just the product of the length and width.

So the best Qi-xiao can do here is to split the dividers as equally as possible.

 It’s a matter here of finding all possible factors of 80 and then to draw up a table giving the dimensions of the crate and then the number of dividers required. So start by noting that 80 = 2⁴ x 5. So the different factorisations of 80 are

2 x 40, 4 x 20, 5 x 16, 8 x 10.

We can now construct our table.

The least number of dividers that we need is 16.

Solution to the Extension:

We deliberately treat these problems using algebra as we expect the more able students to be doing them. However, there is no reason why students shouldn’t use specific values to draw up tables and from there guess the best values. It would be good though, if they could then justify their guess in some way. This justification would vary in ‘depth’ depending on the ability of the student.

 Suppose that we split the dividers with r one way and 2d + 1 – r the other. Then this would give a crate with dimensions r + 1 and 2d + 2 – r. So the number of bottles, B, is given by

B = (r + 1)(2d + 2 – r) = 2d + 2 + (2d + 1)r – r².

The easiest way to do this is to use differentiation. We know that students will not have done this at this stage but show you how to do it and then give an approach that students can use.

dB/dr = 2d + 1 – 2r, so the maximum will occur when 2r = 2d + 1 or r = d + ½. Since r has to be an integer, r = d or d + 1. In these case B = 2d + 2 + (2d + 1)d – d² = d² + 3d + 2, or B = 2d + 2 + (2d + 1)(d + 1) – (d + 1)² = d² + 3d + 2. So the biggest number of bottles that we can put in a crate with an odd number 2d + 1 of dividers is d² + 3d + 2.

Students can do it using a table.

It should be clear that the number of bottles increases to a maximum in the last entry of the table. So the best thing that Qi-xiao can do is to make the crate as close to a square as possible.

Going through the same calculations as above we get a maximum when we have d dividers in each direction and a total of (d + 1)² bottles. In other words Qi-xiao uses a square crate.

We can, of course do this one in the same way that we did the 80 bottles. There will just be that many more cases. However, we will do the problem in its most general form.

By calculus: Let the fixed number of bottles be B and the number of dividers be D. Suppose that r of these dividers go one way and D – r the other. Hence

B = (r + 1)( D – r + 1) = D(r + 1) – (r² – 1), so

D = B/(r + 1) + (r – 1).

Differentiating and putting the derivative equal to zero gives a minimum for D when

r = √B – 1. So D = 2√B – 2. (So the dimensions of the rectangle are both √B – 1.) Now D has to be a whole number so we want it to be as close to 2√B – 2 as possible. This in fact means splitting B into the factors that are as close to the square root of B as we can. (If B is a square then we get a square crate.)

This can be done using a table where we use the factors 1, B; 2, B/2; … √B, √B, for B.

 400 bottles, least dividers?

 2d dividers, most bottles?

 2d + 1dividers, most bottles?

 80 bottles, least dividers?

 11 dividers, most bottles?