Newton's Cows

Explanation

This is a big, messy problem, with an unruly spread of information to be brought under control. And this is exactly its value for the solver: looking at the whole thing, sizing it up somehow, and searching for a good entry point.

The first solution begins with #4 from the list of heuristics, make a table or a chart, which also involves #9, invent notation (a and g for acres and growth). I started with a chart initially to calm myself down and to do something, without any clear conviction that it would lead someplace. Then, once I was into it, I found I was using #13, identifying subgoals, namely, trying to eliminate one of the variables a or g. Then it was #10, writing equations.

The second solution begins with #9, inventing notation, namely e for what a cow eats in a week, then concocting a formula for the total amount of grass eaten under each of the two given conditions, and then plunging into algebra (#10).

I used this with some very able students and simply let them wrestle with it until they got it — which they did. If I had needed to give a hint, I would probably have suggested making a chart.

We could try to find out exactly how much grass one cow will eat in one week. Or we could try to find out how much grass one acre produces in a week (what fraction of its original amount, perhaps). Or we could try to use the information about the two-week situation and the four-week situation to figure out the six-week situation. Is there a formula lurking, maybe the amount of grass eaten as a function of the number of cows and the number of weeks? Or . . .? Lots of possibilities.

Well, let's see. Let’s make a chart about acres and weeks. Let's let a be the original amount of grass on one acre, and let g stand for the amount of grass that grows on one acre in one week. The chart below shows the total amount of grass available on various acreages over different numbers of weeks.

Now that we have the chart, it would be nice if we could get rid of either a or g, that is, express one of them in terms of the other. We might be able to get a formula for how many cows are needed for that 6-acre, 6-week cell in the chart. A worthy subgoal, I hoped.

I tried this: If three cows eat all the grass on two acres plus what grows there in two weeks, how can we compare this to the 2-cow situation? Two cows — that is, 2/3 of three cows — eat 2/3 of the grass on two acres plus 2/3 of what grows on those two acres in two weeks. Using the chart, that's 2/3 of (2a + 4g). And the two cows in four weeks eat twice that much, which is (again from the chart) 2a + 8g. An equation looms.

------------------------ SOLUTION ONE ------------------------------------

		# of weeks
		0	1	2	3	4	5	6
# of acres	1	a	a + g	a + 2g	a + 3g	a + 4g	a + 5g	a + 6g
	2	2a	2a + 2g	2a + 4g	2a + 6g	2a + 8g	2a + 10g	2a + 12g
	3	3a	3a + 3g	3a + 6g	3a + 9g	3a + 12g	3a + 15g	3a + 18g
	4	4a	4a + 4g	4a + 8g	4a + 12g	4a + 16g	4a + 20g	4a + 24g
	5	5a	5a + 5g	5a + 10g	5a + 15g	5a + 20g	5a + 25g	5a + 30g
	6	6a	6a + 6g	6a + 12g	6a + 18g	6a + 24g	6a + 30g	6a + 36g

So here I am: 2 cows in 2 weeks eat 2/3 (2a + 4g) = 4/3 a + 8/3 g.
And 2 cows in 4 weeks (twice as long) eat 2a + 8g, which must be twice as much as 4/3 a + 8/3 g.
So...

I rewrote three crucial cells in the chart, eliminating g.
We know 3 cows in 2 weeks eat 2a + 4g = 2a + a = 3a (how nice! 1 cow in 2 weeks eats 1a.)
And 2 cows in 4 weeks eat 2a + 8g = 2a + 2a = 4a, again giving us 1a per cow per 2 weeks, or 1/2 a per cow per week.
Finally, 6a + 36g = 6a + 9(4g) = 6a + 9a = 15a.
In six weeks, at 1/2 a per cow per week, one cow will eat 3a.
And since there are 15a to be eaten, it will take 5 cows. That's it.

So a humble chart was a good way to begin. Once I had the answer, I tried another approach:

------------------------ SOLUTION TWO ------------------------------------

Suppose we start by thinking of the cows, and what a cow will eat in a week. We can call this e: a cow eats e acres per week.

Then we can say that the total amount of grass eaten under the first of the two given conditions is:
And the second condition gives:
Hmm...
Now, the 6-week conditions ask how many cows will eat in six weeks all the grass on six acres plus all the grass that grows on the six acres in six weeks. Let c represent the number of cows:
Too many variables. But e = 2g so
Can we get a in terms of g so the g's will drop out? Yes, go back to:
So now we have:

So it's 5 cows again. Is this solution better? Is there a still better solution? If this is Isaac Newton's problem, is there a really elegant solution?

Once we know that a = 4g and a cow eats half an acre of grass in a week, it's easy to modify the chart so that it's all a's. Then we can find out which number of acres, for how many weeks, takes an exact number of cows to deal with it. For instance, 3 acres in 4 weeks produces 3a + 12g, or 6a. How many cows chomping away for 4 weeks will consume 6a? One cow, at ½ a per week, will consume 2a, so it will take three cows to consume 6a. On the other hand, 3 acres for 3 weeks produces 3a + 9g, or 3a + 2¼a or 5¼a. We won't get a whole number of cows to consume this in exactly 3 weeks. Here is a question: For every cell that produces a whole number of a's (no fractions), will there be an exact whole number of cows to consume those a's?

The famous hen-and-a-half lays an egg-and-a-half in a day-and-a-half problem (2420.11) and essentially the same problem using variables (2420.12).

The solutions above discuss this problem in terms of my own struggle with it, rather than what I have done with students. I have not found a truly elegant solution, but I suspect there is one — after all, Newton is Newton.