The Chicken Run

Problem

Henry wants to make a rectangular chicken run at the back of the house. He bought 12 metres of fencing wire. What is the largest area run that he can make for his chickens?

What if Henry had 60 metres of fencing?

Teaching sequence

1. Introduce the problem with a piece of string 12 m long. Ask 4 students to make a rectangle with the string. Discuss its perimeter and how they would calculate its area. Ask another 4 students to make another rectangle – estimate whether the area would be greater or smaller that the first one.
2. Read the problem to the class.
3. Brainstorm ways to solve the problem eg. use equipment (squared paper, cm cubes)
4. As the students work ask questions to focus their thinking on working systematically and looking for the relationship between the perimeter and area.
   What can you tell me about that rectangle?
   How are you keeping track of the rectangles that you are making?
   How do you know that you have found all the possible rectangles?
5. Share findings. Encourage the students to make statements about the area and perimeter of the shapes that they have found. Discuss ways to record the measurements to make it easier to see the relationship (table).
6. Pose the second part of the problem.
7. Before they begin working on the problem, remind them to think about ways of working systematically. Check that they planning how they will record their findings.
8. Share solutions.
9. Encourage the students to make statements about the relationship between perimeter and the rectangle with the largest area.

Extensions to the problem

Repeat the problem above with 14 metres of fencing.

Other contexts for the problem
Carpet, layout for a model rainway, pig-pen

Solution to the problem

At this Level we would expect students to tackle this problem by working out the area of a number of rectangles and then choosing the dimensions that give the biggest answer.


Suppose that Henry decided to have a run that was a rectangular shape with the smallest side being 1 metre long. He now has to work out the dimension of the other side to give a value of 5 metres. The area of this run is 1 x 5 = 5 square metres.

But Henry might have used a side length of 2 metres. In that case his chicken run would have had area 2 x 4 = 8 square metres. Or he might have used a side length of 3 metres to give an area of 3 x 3 = 9 square metres.

Now if Henry moves up to 4 metres along the smallest side he’ll find that the longest side is only 2 metres long. There are two things to notice about this. First the smaller side is longer than the longer side and second, Henry has already dealt with the 2 x 4 rectangle. So Henry realises that he has covered all cases for rectangles with whole number side lengths. Hence the maximum area is 9 square metres.

Second part of problem: Henry has to do considerably more calculations when he has more fencing. It is therefore a good idea to summarise the calculations in table form.

From the table we see that Henry’s largest chicken run has area 225 square metres.

Extension:

Some students at this stage might notice that the biggest area in the two examples above occurs when the rectangle is a square. Is this always the case? So let Henry try the 14 metre fencing option.

Henry’s table now uses decimal side lengths and adds in the square case as well. Here the square is again the best option.

This might raise the question in students' minds about using decimal lengths. Perhaps using decimals instead of whole numbers gives a bigger area. At this stage it is worth extending Henry’s first table using decimal lengths. Surprisingly the square always turns out to have the biggest area. The reason behind this requires more maths than students of this Level know but it can be proved at Level 7 or 8.