The Old Chicken Run Problem
use algebraic equations to determine the maximum area of a rectangle with a given partial perimeter.
To be able to do this problem students need to be able to measure lengths and calculate areas of rectangles using the formula: area = length x width. They should probably have tried Peter’s Second String, Level 5 and seen how to use tables to solve the problem or tried Peter’s Third String, Level 6 and seen how to use algebra to solve the problem.
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.
There are seven problems in the Problem Solving section that focus on the perimeter-area relationship. These come in two sets. First there is the set 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 set 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 is enclosed by a wall and a fence? 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 problem and the Extension here are useful as introductions to the type of problem that occurs often in Levels 7 and 8. Here calculus will be used to maximise or minimise various quantities. The method of finding two equations and eliminating one of the variables is one that is often used in max and min problems in calculus.
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.
The farmer was putting a new chicken run up against a brick wall. He had 20 metres of wire to put round the run. If he made a rectangular run, how big an area could he enclose?
- Introduce the problem to the class. Get them to consider how they would approach the problem.
- Initially, let them investigate the problem in any way that they want. They might want to start off with string and make a model of the situation. To use a table they will have to find some equations. They may need some help to do this.
- 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 class 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.
- Get the more able students to try the Extension Problem.
Extension to the problem
The farmer decided that he wanted to have some ‘rooms’ in the chicken run to separate some of the hens. So he used the 20 m of wire slightly differently. We show this in the diagram. What is the biggest area that he can contain now?
We will do this problem by the most sophisticated way open to the students at this point of time. In the diagram below, put in a variable x, the distance that the run is from the wall, and y the length of the run.
Then we can set up some equations. First we know that the length of the chicken wire is 20 m. But it is also equal to x + y + x. So 2x + y = 20 … (1).
Then we know that the area, A, of the chicken run is xy. So A = xy ... (2)
Eliminating y from (1) and (2) gives A = x(20 – 2x). At this point we are in exactly the same situation as we were in Peter's Third String, Level 6. We have a parabola. Its maximum point is halfway between x = 0 and x = 10 (where 20 – 2x = 0). So the maximum is at x = 5. When x = 5, A = 50. So the maximum area is 50 m2.
Note: 1. This problem can be solved using a table as in Peter’s Second String, Level 5. It can also be solved using Calculus but that seems to be an unnecessarily complicated way to solve it.
2. The answer to this problem is not a square. The chicken run of maximum area does not have x = y.
Solution to the extension
The two equations we get this time are 4x + y = 20 and A = xy. Eliminating y now gives the equation A = x(20 – 4x). This parabola has its maximum point halfway between x = 0 and x = 5. So the maximum is at x = 2.5, where A = 25 m2.