Find and use patterns to solve a problem (algebraically or by using a table)
This problem, for which there is more than one solution, follows the problem Square Milk Crates.
In this problem students find patterns and seek the best answer in each situation. The problem therefore belongs in a category involving maximum and minimum problems. These led to the development of calculus, which students may begin in Year 12 when they explore differentiation.
Qi-xiao works with a company that uses square-bottomed milk crates. He wonders if he might get more cartons into rectangular crates using dividers to partition the crates. Using 3 dividers Qi-xiao can fit 4 or 6 cartons into a crate, depending how he uses them. See his crates in the picture.
What is the largest number of cartons that can fit into a rectangular crate if 11 dividers are used.
What is the smallest number of dividers that are needed for a rectangular crate that can hold 80 milk cartons.
- 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 cartons that could be fitted into a crate with 11 dividers? Explain.
- Pose the rest of the problem. Check that it is well understood.
- 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?
- Encourage the students to write down any connections they have found in words and link this to a possible algebraic form.
- Share solutions and reasoning for each part separately. Consider the range of approaches used.
Does your answer make sense? Explain why.
Extension to the problem
What is the largest number of cartons that can fit into a rectangular crate if 2d + 1 or 2d dividers are used? In each case, d is a fixed number.
What is the best-shaped crate that can hold 400 milk cartons? Generalise.
Possible solutions include using a table and using algebra.
A table is used here and an algebraic method is given in the Extension solution.
There are many ways to use the dividers. These are shown in a table. The ‘dividers’ column show the sums that add to 11. The length and width of each possible crate are one more than the addends in the dividers column. The number of cartons is then the product of the length and width.
1 + 10
2 + 9
3 + 8
4 + 7
5 + 6
The best solution is to split the dividers as equally as possible.
To find all possible factors of 80, draw up a table giving the dimensions of the crate and the number of dividers required.
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.
Construct the table.
1 + 39 = 40
3 + 19 = 22
4 + 15 = 19
7 + 9 = 16
The least number of dividers needed is 16.
Solution to the Extension:
In these problems algebra is used. Students may use specific values to draw up tables and from there guess the best values, but having them justify their guess in some way is important. This justification may vary in ‘depth’ depending on 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 cartons, C, is given by
C = (r + 1)(2d + 2 – r) = 2d + 2 + (2d + 1)r – r2.
The easiest way to do this is to use differentiation. Although students will not have done this, it is shown here to give an approach that students can use.
dC/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 C = 2d + 2 + (2d + 1)d – d2 = d2 + 3d + 2, or C = 2d + 2 + (2d + 1)(d + 1) – (d + 1)2 = d2 + 3d + 2. So the biggest number of cartons 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.
1 + 2d
2d + 1
4d + 2
2 + (2d – 1)
3 + (2d – 2)
2d - 2
8d - 8
(d – 2) + (d + 3)
d - 1
d + 4
d2 + 3d - 4
(d – 1) + (d + 2)
d + 3
d2 + 3d
d + (d + 1)
d + 1
d + 2
d2 + 3d + 2
It should be clear that the number of cartons 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 cartons. In other words Qi-xiao uses a square crate.
This can be done in the same way as for the 80 cartons. There will just be that many more cases. However, the problem is done in its most general form.
By calculus: Let the fixed number of cartons be C and the number of dividers be D. Suppose that r of these dividers go one way and D – r the other. Hence
C = (r + 1)( D – r + 1) = D(r + 1) – (r2 – 1), so
D = C/(r + 1) + (r – 1).
Differentiating and putting the derivative equal to zero gives a minimum for D when
r = √C – 1. So D = 2√C – 2. (So the dimensions of the rectangle are both √C – 1.) Now D has to be a whole number so we want it to be as close to 2√C – 2 as possible. This in fact means splitting C into the factors that are as close to the square root of C as we can. (If C is a square then we get a square crate.)
This can be done using a table where we use the factors 1, C; 2, C/2; … √C, √C, for C.
400 cartons, least dividers?
2d dividers, most cartons?
2d + 1dividers, most cartons?
80 cartons, least dividers?
11 dividers, most cartons?