Disagreeing with the Guardian

In this week’s Guardian Weekly (and probably in your Saturday Guardian, but not on-line as far as I can tell) one of the puzzles in Chris Maslanka’s column is as follows:

3 goats will take a week to eat all the grass in a field, and 2 goats will take 2 weeks. How long will 1 goat take?

I do not agree with his answer, which is:

Suppose there is f grass, it grows at a rate of r per week in some appropriate units of consumption and a goat can eat g grass in a week. Then the first two bits of information tell us that -3g + r + f = 0 and -4g + 2r + f = 0. Subtracting the first equation from the second gives -g + r = 0. This means that a single goat eats at the rate the grass grows, so it could never clear the field alone.

But in fact the rate of eating of the grass is not r; it depends on how much grass is left to grow in the field – from the terms of the question, once the goats have eaten it, no more grows, right? So if x is the proportion of the field still covered in grass, then the rate at which the grass is growing is rx, not r. So my first equation here, where you have n goats and f is the (variable) amount of grass in the field, is

df/dt = rx – ng

x is of course also variable. The goats are presumably not eating at a constant rate of so many square metres per day, but rather eating at a rate of so many kilos or bushels of grass per day. It’s fairly easy to show that the rate of clearance of the field, dx/dt, is inversely proportional to how much the grass has grown in the meantime, in other words to the variable f (whose initial value, the amount of grass in the field at the start, is f0):

dx/dt = -ng / f

Hah, this looks familiar from undergraduate maths. If I divide one equation by the other, then

df/dx = f – rxf / ng = (1 – rx/ng) f

which means, after a bit of scrabbling with my memories of differential equations, that

f = f0 (e x – rx2 / 2ng – 1) / (e1 – r/2ng – 1)

OK, so it’s a dubious exponential quadratic; the grass is at its longest when the goats have cleared ng/r of the total area, but as for when they have eaten the lot, that takes considerable number-crunching. I’ve long forgotten how to programme a computer to do this, but I bet just looking at it that if you can get r and g right to give the correct numbers for 1 and 2 goats, the goat on his own will take quite a lot longer to eat his way through the grass, but will get there in the end.

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