Peter keeps a piece of string from a parcel that came for his birthday. It is 30 cm long.
He plays with it and makes different shapes.
He thinks that the rectangle with the biggest area that he can make, is a square.
His sister Miri disagrees.
Who is right and why?
In this problem students measure lengths, calculate the perimeters and areas of rectangles using the formulae: perimeter = twice length plus twice width and area = length x width. Students should know how to use a table as they investigate the mathematical question that this problem probes, what can we say about the rectangle of biggest area that has a fixed perimeter?
We find that the largest area comes from a square and the smallest area is as small as we like to make it.
Related problems are in two sets.
The first includes: Peters’ String, Measurement, Level 4, Peters’ Second String, Measurement, Level 5, Peters’ Third String, Algebra, Level 6, The Old Chicken Run Problem, Algebra, Level 6 and the Polygonal String Problem, Algebra, Level 6.
The second includes: Karen’s Tiles, Measurement, Level 5 and Karen’s Second Tiles, Algebra, Level 6
Copymaster of the problem (English)
Copymaster of the problem (Māori)
squared paper or graph paper
pieces of string of various lengths
ruler
Problem
Teaching sequence
 Introduce the problem to the class and discuss with the students approaches they might use.
 Give each pair of students a piece of string. Perhaps give different groups different lengths. Have students investigate Peter’s conjecture in any way that they want. At some stage though they will probably record some equations and may need help.
 As they work, check on their progress, and encourage systematic recording.
If they have an approach that doesn’t need a table, have them follow it through to see if it will work, as there are many ways to solve this problem.  Pose the Extension Problem when appropriate. Students should see that they can use the original problem in the solution.
Share ideas if some students are struggling.  Have students share solutions in a class discussion, explaining their strategies and methods of recording.
Extension problem
Peter’s string is 30 cm long and Miri’s is 20 cm long. We know that Miri can make some rectangles that have the same area as Peter (see Peter’s String, Level 4). Presumably she can’t make all the areas that Peter can. Presumably he can make all of the areas that she can. Can you justify these two hunches?
Solution
One way to do this is to use a table (as in Peter’s String, Level 4.) First we have to set up the table.
What do we know? If we make Peter’s string into a rectangle with side lengths L and W, then 2L + 2W = 30 or L + W = 15.
We know that LW = A, where A is the area of the rectangle. But we don’t know what A is. We want to find the maximum value it can have subject to L + W = 15. Draw up a table with L + W = 15 and calculate LW for various values of L to see where the maximum value of A is.
L

W

L + W

A = LW

1

14

15

14

2

13

15

26

3

12

15

36

4

11

15

44

5

10

15

50

6

9

15

54

7

8

15

56

8

7

15

56

9

6

15

54

10

5

15

50

11

4

15

44

Table 1
This seems to give a peak at L = 7 or 8. But there may be a bigger value of A for a value of L between these two values. Draw up another table.
L

W

L + W

A = LW

7.1

7.9

15

56.09

7.2

7.8

15

56.16

7.3

7.7

15

56.21

7.4

7.6

15

56.24

7.5

7.5

15

56.25

7.6

7.4

15

56.24

7.7

7.3

15

56.21

7.8

7.2

15

56.16

7.9

7.1

15

56.09

Table 2
Table 2 suggests that the top value for A is 56.25 when L = W = 7.5. It would be a good idea to check for L in between 7.4 and 7.5 to make sure that we don’t get an unexpected peak there. Have your students check this.
The conclusion is that indeed the maximum area is when L = W = 7.5. Hopefully your students will have used a variety of different string lengths and they will have all come to the conclusion that the rectangle of maximum area with fixed perimeter, is a square.
Note: The proof above leaves one point in doubt. That point is the implicit assumption that A gradually increases as L increases up to a certain point. Then as L increases further A decreases gradually. The tables suggest that this is what is happening (and it actually is), but we may have taken the different values of L too far apart and as a result have missed some jumps in the values of A. The proof in Peter’s Third String, Level 6, avoids this problem by appealing to the properties of a parabola.)
Extension Solution:
Consider the smallest area Peter can make. Note from our experience with the tables, as we change the dimensions of the rectangle and make the shortest side shorter, we change the area of the rectangle. We begin to suspect that we can make rectangles of smaller and smaller area. But how small can we go? What if Peter had one side of length 0.01 cm? Then the other would be 14.99 cm. This would give an area of 0.14 cm^{2}. Could get below that by taking one side of length 0.001 cm (area here is 0.014999 cm^{2}). Then we could go smaller again if one side was 0.0001 cm, and so on. It should be starting to become clear that, by choosing one side small enough, we could get the area of our rectangle smaller than any chosen small number. This means that Peter’s rectangle can be made to go as close to zero as he likes (or is practical).
The same thing is clearly true for Miri.
We know from the first part of this problem, that Peter and Miri’s strings give them a maximum area when the rectangle they form is a square. In Peter’s case this square has side length 7.5 cm. So he can produce an area of 56.25 cm^{2}. Miri’s square has side length 5 cm to give an area of 25 cm^{2}. Peter and Miri can both make their areas go from their maximum down to zero.
As a result we know that Miri cannot produce all the areas that Peter can but that he can produce all the areas that she can.
Note: This problem can also be solved using a graphics calculator, a computer programme, a graph or calculus. A formal noncalculus proof is given in Peter’s Third String, Level 6.