# Report on fortnightly maths prize, 25 November 2016

A goat is tethered to the outside of a square barn in the middle of a large field. Where on the barn should the tether be attached to give the goat the biggest grazing area?

Prizes were won by Conner Lake and Bolaji Atanda. Mention should also go to Rashi Bhatt and Genie Louis de Canonville, who worked on it and made some progress but in the end didn’t submit an entry.

If the tether is short compared to the side of the barn, then with the tether attached at a corner the goat gets three 90-degree sectors of a circle of radius equal to the length of the tether. Anywhere else the goat gets only two 90-degree sectors like that, plus maybe (or maybe not) a third sector of smaller radius. So a corner is best.

Trickier is showing that corner is still best even if the tether is longer.

You can calculate particular cases (e.g. tether exactly equal to circumference of barn). You can calculate the case where the tether is very long compared to the side of the barn. Both still show the goat gets a better deal with the tether attached at a corner.

But how can we prove that is true whatever the length of the tether?

If the tether’s length is less than or equal to the circumference of the barn, then there is a simple argument from symmetry. Let F(x) = area in terms of the distance x of the tether from the midpoint of a side, taking barn-side-length 2b and tether-length t as constants. Symmetry tells us that F(x) is the same for every midpoint; and F is symmetric around x=0.

And more. F(x) will be a sum of terms each of which is the area of a 90-degree sector of a circle of radius t, or (t-b+x), or (t-b-x), or (t-3b-x), or (t-3b+x). It will be a sum of quadratic functions of x, in each of which term x2 has a positive coefficient. Its graph will look like a symmetric stitching-together of bits of vertex-at-the-bottom parabolas. So it has a minimum at x=0, and its greatest value at the greatest and least possible values of x, namely x=b or x=−b, i.e. at a corner.

That argument breaks down if the tether’s length is greater than the circumference of the barn, because then F(x) is a more complicated function.

This is the simplest argument I can find to prove the conclusion for all lengths of tether.

Suppose x increased by a small amount δx. Then

δF = 0 [for the grazing area above the broken horizontal line in the diagram; that doesn’t change]

+ δx.(perimeter on the right below the broken line, clockwise to the cusp A if the tether is long enough to wind round the barn, or to the furthest clockwise stretch round the barn wall if the tether is shorter)

− δx.(perimeter on the left below the broken line, anticlockwise to the cusp A or to the furthest anticlockwise stretch)

+ terms of order (δx)2.

But the perimeter on the right > perimeter on left. So dF/dx≥0 (for x≥0; and dF/dx≤0 for x≤0). Therefore F has a maximum at a corner (x=b or x=−b). 