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?
Solution
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 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.
This problem can also be solved using a table. It can also be solved using Calculus but that seems to be an unnecessarily complicated way to solve it.
Extension
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?
