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
Note that this is the same problem as we used in Peter’s Second String, Level 5 but at this level we expect students to have a more sophisticated way to solve the problem.
The extension problem here is quite difficult. If the students can make progress with any quadrilateral, then they have done well. The method shown here may not be the easiest way to solve the problem. If your class comes up with a better approach we would like to see it so that we can add it to this problem.
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 help if they have tried Peter’s Second String, Level 5 and have 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 web site 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.
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.
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?
- Introduce the problem to the class. Get them to consider how they would approach the problem.
- Let them investigate Peter’s conjecture in groups using any approach that they want. (Although you might decide to move them in a more sophisticated direction.) 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. Encourage them to set up some equations and reduce the number of dependent variables to one.
- 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.
- Encourage the more able students to try the Extension Problem. You may want to give them a few days to think about it.
Extension to the problem
If Peter made any quadrilateral shape with his string, would the maximum area he could get be produced by a square?
We have already done this problem using a table in Peter’s Second String, Level 5. But at this level students should begin to see how to use algebra more effectively in such problems so we would encourage you to move them in this direction here.
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. This gives us W = 15 – L.
Then we know that LW = A, where A is the area of the rectangle. So we can substitute for W into this equation and at least arrive at an expression for A that only involves the one variable L. This expression is A = L(15 – L). Now we can graph this function because it’s a parabola. And we know from past experience that (i) it is has a maximum value because the coefficient of L2 is negative; and (ii) it crosses the L-axis when L = 0 and L = 15.
But parabolas have their axis of symmetry and maximum point, midway between the points where they cross the L-axis. So the axis of symmetry of the parabola above is at L = 7.5. This is also where its maximum point is.
But if L = 7.5, then W = 15 – L = 15 – 7.5 = 7.5. So W = L and the rectangle has to be a square.
Solution to the extension
We do this in easy steps. First the rectangle of perimeter 30 cm with biggest area is a square. This has been done in the main problem. Second, a parallelogram with no internal angles of 90 has a smaller area than a rectangle with the same side lengths. To see this, consider the parallelogram and the rectangle below. Clearly the rectangle has area bc. The parallelogram has area base x height. Since its height is less than c, its area is less than bc. So the parallelogram has a smaller area than the rectangle even though they both have the same perimeters.
Third, a quadrilateral with an exterior angle less than 180° has a smaller area than a corresponding one with all exterior angles bigger than 180°. To see this, look at the diagram below. Clearly the quadrilateral on the right has the largest area yet they both have the same perimeters.
Fourth, the quadrilateral on the right above has smaller area than a quadrilateral whose diagonals are perpendicular. To see this, imagine that the points A and B are fixed and that A to C to B is a piece of string.Place a pencil inside the string at C. Move the pencil keeping AC and BC straight (taut). What is the biggest area of the triangle ABC? Well as C moves, the base AB remains the same. So the biggest area is found when the height of the triangle is the largest. This occurs when the perpendicular from C to AB goes through the midpoint of AB. Call this position for C, K. (Can you see that AK = KB?)
Notice that the quadrilateral formed by AKBD has the same perimeter as ACBD but AKBD has the bigger area.
We can do exactly the same thing with the point D. The position of D which makes triangle ABD have the biggest area is when D is above the midpoint of AB. Call this point L. Then the quadrilateral AKBL has the same perimeter as ABCD but it has a bigger area. What’s more, KL is perpendicular to AB.
Fifth, we want to show that among all quadrilaterals with the same perimeter and with perpendicular diagonals, the one with the biggest area is a parallelogram. Consider the quadrilateral below.
By the argument of ‘Four’, we can assume that AB = AD and that BC = CD. Now repeat the argument of ‘Four’ using B as the point to put the pencil and A and C as the fixed points. The argument we used above then shows that we can move B to a place above the midpoint of AC and in the process increase the area of triangle ABC. Repeating this on triangle ACD, we see that the area of this triangle is maximised when D is above the midpoint of AC. So we have a quadrilateral whose area is bigger than the original quadrilateral ABCD.
In this new quadrilateral, AB = BC = CD = DA. From here it is easy to show that opposite angles are equal and so the quadrilateral is a parallelogram (with equal sides – a rhombus).
Now lets recap. We can show that among all rectangles with a given perimeter, the one with biggest area is a square. Then any other quadrilateral can be changed into another quadrilateral with the same perimeter and bigger area, using ‘Three’ and/or the ‘pencil’ approach of ‘Four’. What’s more this quadrilateral has to be a parallelogram. But we know that for every parallelogram with a given perimeter there is a rectangle with the same perimeter and a larger area. Hence among all quadrilaterals with a given perimeter, the square is the one with the biggest area.