A Sheep-pen Problem

Problem statement:

A farmer has 60 identical hurdles, with which she plans to build a sheep-pen. She wants the area of the pen to be as large as possible. Two hurdles can be fixed together end-to-end to form a straight line, or to be at right angles; no other angles are possible.

Where she plans to build the pen, there's a long straight sheep-proof fence. Whenever the end of a hurdle is next to the fence, she can make a sheep-proof join. She realises that she can use the fence to form one side of the pen.

What shape should she make the pen?

You may want to try to solve the problem, before you look at the two solutions below.

Solution using calculus: show

Solution using symmetry: show

Comparison of solutions: show

If you're still not convinced that symmetry can be easier to use than calculus, try this puzzle:

As above, but instead of 60 short hurdles, she has five equally long hurdles, which can be fixed together end-to-end at any angle. She can still use the existing long fence. What shape should she make the pen?

If you've understood the symmetry-based solution to the previous problem, you should find this one easy using symmetry. It's also possible using calculus.

This is one of several pages on using symmetry in mathematics.