Farmer John kept his new lambs in an enclosure surrounded by eight long hurdles arranged as in the diagram below. One year, after a particularly good season, Farmer John found that he had 48 lambs instead of the usual 30.

‘If only I had two more hurdles,’ he sighed to his wife Mwyn, ‘I could double the size of the enclosure.’

But unfortunately he had thrown out anything that might have come in useful in a recent spring-clean.

Perhaps you can help Farmer John to rearrange the eight hurdles to form an enclosure large enough to take his new flock?

Hint 1

What figure of given perimeter has the greatest area?

Hint 2

If the hurdles are 1 unit long, what area do we require for the new enclosure?

Solution

Taking the hurdles to have unit length, the given enclosure has area 3. This works out at 1 square unit for each 10 lambs. Thus we are looking for a new enclosure having area just under 5.

The most obvious way to arrange the hurdles is to place them in the shape of a 2 x 2 square. However, this will only have area 4.

Some or all of the following factors lead us to the solution.

(1) Experimenting, we find that there are only a small number of ways of arranging eight hurdles in ‘horizontal’ and ‘vertical’ positions. This suggests that some diagonal placements might be useful.

(2) Amongst all shapes of given perimeter the circle has the largest area. This is the isoperimetric problem.

(3) There are eight hurdles.

We therefore try the regular octagon as a likely solution. We see that the area of this octagon has area 5, so the enclosure is large enough for the 48 lambs with a little room to spare.

Extensions

1. Devise a similar problem in which Farmer John has just six hurdles.

2. What would be the situation if the lambs were surrounded by a flexible wire of given length, supported by as many posts as required?

3. We can vary Extension 2 by assuming that a long straight shed wall forms part of the boundary. Make up some further variations of your own.

Hint 1

Hint 2

Solution

Extensions