Peter's Third String

The 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

1. Introduce the problem to the class. Get them to consider how they would approach the problem.
2. 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.
3. 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.
4. If a lot of the pairs are having problems, then you may want a brainstorming session to help them along.
5. Share the students’ answers. Get them to write up their work in their books. Make sure that they have carefully explained their arguments.
6. 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?

Solution

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.