use algebraic equations to determine the maximum area of a rectangle with a given partial perimeter.
This problem challenges students to maximise an area given a number of parameters. The method of finding two equations and eliminating one of the variables is one that is often used in maximum and minimum problems in calculus.
To solve this problem students must be able to measure lengths and calculate areas of rectangles using the formula: area = length x width, and to apply an elementary knowledge of parabolas. It would also be an advantage if they have used a table and applied algebra to find solutions to the problems: Peter’s Second String, Level 5 and Peter’s Third String, Level 6.
Mathematics is more than doing calculations or following routine instructions. Thinking and creating are at the heart of the subject. Though some problems have a set procedure or a formula that can be used to solve them, the most worthwhile problems require the use of known mathematics in a novel way.
There are seven problems related to perimeter and area.
The first set includes: Peters’ String, Measurement, Level 4, Peters’ Second String, Measurement, Level 5, Peters’ Third String and Polygonal String Problem Algebra, Level 6. These show the non-link between rectangles’ areas and perimeters, including showing that among all quadrilaterals with a fixed perimeter, the square has the largest area. The final problemexplores the areas of regular polygons with a fixed perimeter and shows that they are ‘bounded above’ by the circle with the same perimeter.
The farmer is putting a new chicken run up against a brick wall. He has 20 metres of wire to put round the run. If he makes a rectangular run, how big an area can he enclose?
The farmer wants to have some ‘rooms’ in the chicken run to separate some of the hens. So he uses the 20 m of wire slightly differently.
See the diagram. What is the biggest area that he can contain now?
This solution is the most sophisticated available to the students at this point of time.
In the diagram below, the variable x is the distance that the run is from the wall, and y the length of the run.
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.
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.
Printed from https://nzmaths.co.nz/resource/old-chicken-run-problem at 5:53pm on the 20th January 2021