There is more than one solution to this problem which focuses on finding patterns.
Possible strategies that students may use to find a solution include solving a simpler problem, using a table and using algebra.
Qi-xiao works with a company that uses square-bottomed milk crates. A square number of milk cartons can fit in a crate using dividers. Using 2 dividers Qi-xiao can fit 4 cartons into a crate. Using 4 dividers he can fit 9 cartons in another sized crate. Both of these situations are shown in the picture.
How many cartons can fit into a square crate if 30 dividers are used?
How many dividers are needed for a square crate that can hold 100 milk cartons?
- Show the students the picture of a carton with 4 dividers and ask:
What mathematical questions could we pose about this carton?
- Pose the problem for the students to solve, and ask for ideas on how they might get started..
- As the students work on the problem ask questions that encourage them to think about their problem solving strategies:
What strategy are you using to solve the problem? Are you making progress with it?
How are you keeping track of what you do?
Have you seen similar problems before?
- Also ask questions that encourage the students to look for and explain patterns in the number of dividers.
What have you found out about the dividers?
Does that help you solve the problem? How?
What patterns can you see in the problem?
What helps you find patterns in problems like these? (tables)
- Share solutions and justifications.
Extension to the problem
Using the square crates, Qi-xiao decides to work out some formulae.
If he has 2d dividers, what is the maximum number of cartons that he can put in the crate?
If he wants to fit c2 cartons in a square crate, how many dividers does he need?
There are a number of possible solutions to these problems. They include solving a simpler problem using a table and using algebra. Both are given below.
Option 1: The 30 dividers problem.
Using a table and considering simpler cases first, it becomes clear that each carton must fit into a square hole formed by the dividers.
That means that an equal number of dividers have to be used in each direction. What’s more there is always one more square per side than the number of dividers perpendicular to that side.
Numbers of divider
Number of milk cartons
The table can be continued up to 30 if the pattern is not obvious to students from the start.
The 100 milk cartons problem.
One way to do this is to work backwards in the table. Since 100 = 102, and 10 – 1 = 9, then 2 x 9 is the number of dividers that have to be used. So Qi-xiao will have to find 18 dividers.
Solution to the extension
The square numbers become evident and need to be linked algebraically with the number of dividers. The number of dividers occurs in pairs. If each pair is considered as a single case then the table becomes.
No. of dividers
No. of milk cartons
(d + 1)2
Focusing on the number of dividers 2d we can generate a quadratic formula for the number of milk cartons.
So the number of cartons = (d + 1)2.
If we have c2 cartons, then we can work from the right to the left in the table.
Corresponding to c2 we have 2(c – 1) in the middle column. So
the number of dividers = 2(c – 1).
Option 2: The 30 dividers problem.
By drawing a diagram it can be seen that half the number of dividers are used widthways and half lengthways. So 15 dividers will be used on one side of the crate.
Hence 15 divides on one side will give 16 carton spaces. Thus the total number of carton spaces in the square crate is 16 x 16 = 256.
The 100 milk cartons problem.
100 cartons mean 10 cartons on each side so there are 9 dividers on each side. Hence there are 18 dividers altogether.