Karen's Second Tiles
Determine the maximum area of a regular polygon with a given perimeter
Appreciate the concept of limit as it applies to the area of regular n-gons and circle that both have the same perimeter
To be able to do this problem students need to be able to measure lengths and calculate areas of triangles using the formula: area = ½ base length x height. They should probably have tried Peter’s Third String, Level 6 and seen how to use algebraic equations to solve problems.
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 by following our mathematical nose.
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. Throughout this web site we are hoping to motivate students to think about what they are doing and see connections between various aspects of this. The mathematical question is again an optimisation problem. These are typical of a lot of mathematics. Indeed differential calculus, that comes up at Levels 7 and 8 if not before, is very good at solving a whole range of such questions. In fact a lot of the algebraic work here is a good introduction to the work that is done at the next two Levels.
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 encircle. 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 of a rectangle, we then use in a completely different application – 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.
But having spent time with rectangles, the obvious thing to do is to look at other shapes. We actually look at polygons and their relation with circles but there is no reason why you shouldn’t look at triangles or hexagons. Here you might ask whether you can find two triangles with the same area and perimeter or what is the triangle with given area that has maximum perimeter. We have actually avoided these last two questions because of the difficulty of the maths that would be required to solve them. However, we may have got it wrong. There may be some nice answers that are relatively easy to find. If so, please let us know.
The Problem
Karen discovered some old tiles that had been left at a building site and she started to makedifferent rectangles with them. She got to wondering what were the biggest and the smallest perimeters that she could find, for rectangles with areas equal to 100 cm2.
Teaching sequence
- Introduce the problem to the class. Get them to consider how they would approach the problem.
- Let the class investigate the problem in any way that they want. At some stage though they will probably have to write down some diagrams and equations. They may need some help at this point.
- Move around the groups as they work to check on progress. Encourage them to draw large diagrams to show clearly what is going on.
- If a lot of the pairs are having problems, then you may want a brainstorming session to help them along. You may need to remind them of the value of using a table.
- The Extension problem involves some difficult algebra so it may be best tackled by only the more able students. However, it does contain ideas that will help prepare them for work at Levels 7 and 8.
- Share the students’ answers. Get them to write up their work in their books. Make sure that they have carefully explained their arguments.
Extension to the problem
Karen then got to wondering whether it was true that two different rectangles could have the same area and the same perimeter?
Marty didn’t think that two rectangles could have the same area and the same perimeter. But he couldn’t convince Karen of that. Can you?
Solution
This problem can be set up algebraically. From there it can be solved using a table or by drawing a graph.
| Let the rectangles’ variable perimeter be P. If the length of the rectangle is L and the width W (both variables), then | |
| 100 = LW … (1) and P = 2(L + W) … (2). | |
| From equation (1), we can write L in terms of W. | |
| L = 100/W … (3). | |
| Substituting for L in (2) gives us | |
| P = 2(100/W + W) … (4). | |
At this point we can move in at least two directions. These are tables and graphs. We use the table method first.
|
W |
P |
|
5 |
50 |
|
6 |
45.33 |
|
7 |
42.57 |
|
8 |
41 |
|
9 |
40.22 |
|
10 |
40 |
|
11 |
40.18 |
|
12 |
40.67 |
|
13 |
41.38 |
|
14 |
42.29 |
So it looks as if the smallest value for P is 40 when W = 10. We should really check this out by choosing some values closer to 10 (perhaps 9.5, 9.6, 9.7, 9.8, 9.9, 10, 10.1, 10.2, 10.3) to be reasonable sure about this. But this is the actual answer.
Now let’s try the graphical method. This is much easier if your class has graphics calculators (or a computer that will draw graphs) as the machine will do the hard work for them. If not, then your students will have to draw a careful graph on graph paper. No matter what route you choose you should get a graph like the one below.
So as the result of both sets of work above, we can see that if the area of a rectangle is 100 cm2, then the smallest value of the perimeter is 40 cm when W = 10 cm. (You might note that this occurs when the rectangle is a square.)
But what is the biggest value that the perimeter can have? It may be easiest to see this from the graph. Clearly as W approaches 0, P approaches infinity. And as W approaches infinity, P also approaches infinity. So P can be as big as we want it to be. There is in fact no biggest value for P.
This requires a good grasp of algebra. Suppose that we have two rectangles with lengths and widths L, W and A, B, respectively. If they have the same are, then
| LW = AB … (1) | |
| On the other hand, if they have the same perimeter, then | |
| L + W = A + B … (2) | |
| Multiplying equation (2) by W we get | |
| LW + W2 = AW + BW … (3) | |
| Substituting for LW from equation (1) we get | |
| AB + W2 = AW + BW … (4) | |
| Rearranging gives | |
| AB – AW = BW – W2 … (5) | |
| Or | |
| A(B – W) = W(B – W) … (5) | |
| Now this factorises to give | |
|
(A – W)(B – W) = 0 … (6) |
|
| From equation (6) we see that either A = W or B = W. | |
Substituting the first of these into equation (1) gives B = L (and A = W), and so both rectangles are the same or it gives A = L (and B = W) and again both rectangles are the same. So it is not possible for two different rectangles to have both the same area and the same perimeter.
| Attachment | Size |
|---|---|
| KarensSecondTile.pdf | 38.37 KB |
| KarensSecondTileMaori.pdf | 49.29 KB |
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