Peter's Second String
determine the maximum area of a rectangle with a given perimeter
determine the range of areas of a rectangle with a given perimeter
pose questions for mathematical exploration
prove or refute mathematical conjectures
To be able to do this problem students need to be able to measure lengths and calculate the perimeters and areas of rectangles using the formulae: perimeter = twice length plus twice width and area = length x width. In addition, it would probably help if they have tried Peter’s String, Level 4 and seen how to use tables.
Apparently in some areas of New Guinea they measure the area of land by its perimeter. When you think about it this isn’t such a good idea. A piece of land can have a relatively large perimeter and only a small area. This sequence of problems is built up from this simple bad idea.
Seven problems have been spawned by the perimeter-area tangle. These come in two waves. First there is the string of Peter’s String problems. These are 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. These follow through on the non-link between rectangles’ areas and perimeters, going as far as showing that among all quadrilaterals with a fixed perimeter, the square has the largest area. In the second last of these five problems we are able to use an idea that has been developed to look at the old problem of maximising the area of a chicken run. This is often given as an early application of calculus but doesn’t need more than an elementary knowledge of parabolas. The final problem looks at the areas of regular polygons with a fixed perimeter. We show that they are ‘bounded above’ by the circle with the same perimeter.
The second string of lessons looks at the problem from the other side: does area have anything to say about perimeter? This leads to questions about the maximum and minimum perimeters for a given area. The lessons here are Karen’s Tiles, Measurement, Level 5 and Karen’s Second Tiles, Algebra, Level 6.
Mathematics is more than doing calculations or following routine instructions. Thinking and creating are at the heart of the subject. Though there are some problems that have a set procedure or a formula that can be used to solve them, most worthwhile problems require the use of known mathematics (but not necessarily formulae) in a novel way.
Throughout this Problem Solving section of the website we are hoping to motivate students to think about what they are doing and see connections between various aspects of what they are doing. The mathematical question asked here is what can we say about the rectangle of biggest area that has a fixed perimeter? This question is typical of a lot of mathematical ones that attempt to maximise quantities with given restrictions. There are obvious benefits for this type of maximising activity.
The ideas in this sequence of problems further help to develop the student’s concept of mathematics, the thought structure underlying the subject, and the way the subject develops. We start off with a piece of string and use this to realise that there is no direct relation between the perimeter of a rectangle and its area. This leads us to thinking about what areas are possible. A natural consequence of this is to try to find the largest and smallest areas that a given perimeter can encompass. We end up solving both these problems. The largest area comes from a square and the smallest area is as small as we like to make it.
Some of the techniques we have used to produce the largest area are then applied in a completely different situation – the chicken run. This positive offshoot of what is really a very pure piece of mathematics initially, is the kind of thing that frequently happens in maths. Somehow, sanitised bits of mathematics, produced in a pure mathematician’s head, can often be applied to real situations.
The next direction that the problem takes is to turn the original question around. Don’t ask given perimeter what do we know about area, ask given area what do we know about perimeter. Again there seems to be no direct link.
Problem
Peter had kept a piece of string that had been on a parcel that had come for his birthday. It was 30 cm long. He played with it and made different shapes out of it. Then he got stuck on rectangles. He wasn’t sure but he thought that the rectangle with the biggest area that he could make was a square. His sister Veronica said that was crazy but she didn’t have a good reason for saying that. Who was right and why?
Teaching sequence
- Introduce the problem to the class. Get them to consider how they would approach the problem.
- Give each pair of students a piece of string. It’s probably a good idea to give different groups different lengths. Let them investigate Peter’s conjecture in any way that they want. At some stage though they will probably have to write down some equations. They may need some help at this point.
- Move round the groups as they work to check on progress. If they have an approach that doesn’t need a table, then follow it through to see if it will work. (There are many ways to do this problem.)
- The Extension Problem may actually be easier to do than the original but they have to see that they can use the original problem in the solution. Help them talk themselves through the smallest area. Get them to put together a reasonable argument. If a lot of the pairs are having problems, then you may want a brainstorming session to help them along
- Share the students’ answers. Get them to write up their work in their books. Make sure that they have carefully explained their arguments and any parts where they feel that they haven’t quite covered all of the difficulties.
Extension problem
Peter’s string is 30 cm long and Veronica’s is 20 cm long. We know that Veronica 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 in much the same way that we did in Peter’s String, Level 4. But first we have to set up the table. (This problem could also be solved using a graphics calculator, a computer programme, a graph or various other means, even calculus. A formal non-calculus proof is given in Peter’s Third String, Level 6.)
What do we know? Well if we make Peter’s string into a rectangle with side lengths L and W, then 2L + 2W = 30 or L + W = 15.
Then we know that LW = A, where A is the area of the rectangle. But we don’t know what A is. In fact we want to find the maximum value it can have subject to L + W = 15. So we’ll 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. So we now 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 seems to suggest that the top value for A is 56.25 when L = W = 7.5. But 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. We leave your students to check this.
The conclusion though 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: As a proof the one above leaves one point to be desired. 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. Now 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. The first thing to note is that, 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 cm2. But we could get below that by taking one side of length 0.001 cm (area here is 0.014999 cm2). 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 Veronica.
Now we know from the first part of this problem, that Peter and Veronica’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 cm2. On the other hand, Veronica’s square has side length 5 cm to give an area of 25 cm2. Peter and Veronica can both make their areas go from their maximum down to zero.
As a result we know that Veronica cannot produce all the areas that Peter can but that he can produce all the areas that she can.
| Attachment | Size |
|---|---|
| PetersSecond.pdf | 41.09 KB |
| PetersSecondMaori.pdf | 52.48 KB |
Similar Resources
The Old Chicken Run Problem
use algebraic equations to determine the maximum area of a rectangle with a given partial perimeter.
Polygonal Strings
This is a problem from the number and algebra strand.
Peter's Third String
Determine the maximum area of a rectangle with a given perimeter
Consider how the area of a quadrilateral changes as its shape changes
Interpret a relationship from a graph
Peter's String
Determine the area of a rectangle with a given perimeter;
Pose questions for mathematical exploration
Make conjectures in a mathematical context
Prove or refute mathematical conjectures
The Chicken Run
explain the relationship between the area and perimeter of rectangles
use a table to solve a problem
devise and use problem solving strategies to explore situations mathematically (guess and check, be systematic, make a table, make a drawing)
