# The grazing cows of Sir Isaac Newton

Sir Isaac Newton's book "Arithmetica Universalis" contains the following famous puzzle:

In $4$ weeks, $12$ cows graze bare $3\frac13$ acres of pasture land, and in $9$ weeks, $21$ cows graze bare $10$ acres of pasture land. Accounting for the uniform growth rate of grass and assuming equal quantities of grass per acre when the pastures are put into use, how many cows will it take to graze bare $24$ acres of pasture land in a period of $18$ weeks?

Nowadays, some routine knowledge of highschool algebra suffices to find the answer to Sir Isaac's puzzle: It will take $36$ cows.

This puzzle asks you to develop the general solution for general grazing cow puzzles of this type:

Sir Isaac Newton's generalized puzzle: Assume that

• it takes $c_1$ cows to graze bare $a_1$ acres of pasture in $w_1$ weeks;
• it takes $c_2$ cows to graze bare $a_2$ acres of pasture in $w_2$ weeks;
• it takes $c_3$ cows to graze bare $a_3$ acres of pasture in $w_3$ weeks.

What simple algebraic relation does there exist between the nine quantities $c_1,c_2,c_3$ and $a_1,a_2,a_3$ and $w_1,w_2,w_3$?

The answer can either be given as an ugly equation (by plodding and laborious work), or in terms of a simple and elegant determinant (by mathematical insight).

First things first: we will declare some constants:

• $\text{COW_GRASS_PER_WEEK}=$ quantity of grass eaten by a cow in one week
• $\text{ACRE_GRASS_PER_WEEK}=$ quantity of grass that grows in one acre in one week
• $\text{INITIAL_GRASS_PER_ACRE}=$ initial quantity of grass per acre

We will call them respectively $CGW$, $AGW$ and $IGA$.

The variables are going to be the same as in the question: $c_1,c_2,c_3$ for the number of cow, $a_1,a_2,a_3$ for the number of acres and $w_1,w_2,w_3$ for the number of weeks.

To resolve the example problem we have:

• $c_1=12$, $a_1=\frac{10}{3}$,$w_1=4$
• $c_2=21$, $a_2=10$, $w_2=9$
• $c_3=X$, $a_3=24$, $w_3=18$

We want the amount of grass eaten by the cows to be the same as the amount of grass that grew in a fixed time period plus the initial amount, so that we have the generalized equation:

$$c\cdot w\cdot CGW=a\cdot w\cdot AGW + a\cdot IGA$$

In our case:

• $12\cdot4\cdot CGW=\frac{10}{3}\cdot4\cdot AGW + \frac{10}{3}\cdot IGA$,
• $21\cdot9\cdot CGW=10\cdot9\cdot AGW + 10\cdot IGA$ and
• $X\cdot18\cdot CGW=24\cdot18\cdot AGW + 24\cdot IGA$

which leads us to $CGW=\frac{10}{9}AGW$ and $IGA=12\cdot AGW$. After substituting we get $20\cdot X \cdot AGW = 720 \cdot AGW$, which finally implies the answer $X=36$.

Now how can we generalize this? Here is where matrices come into play! We have 3 equations and 3 variables of the form: $$c_i\cdot w_i\cdot CGW=a_i\cdot w_i\cdot AGW + a_i\cdot IGA$$ respectively $$a_i\cdot w_i\cdot AGW + a_i\cdot IGA - c_i\cdot w_i\cdot CGW = 0.$$ The resulting linear equation system has the following coefficient matrix: $$\begin{bmatrix}a_1\cdot w_1 & a_1 & -c_1\cdot w_1\\a_2\cdot w_2 & a_2 & -c_2\cdot w_2\\a_3\cdot w_3 & a_3 & -c_3\cdot w_3\end{bmatrix}.$$
If the determinant of this matrix was nonzero, then the matrix would be invertible and we would have a unique solution for $CGW$, $AGW$, and $IGA$. But the solution isn't unique, as we may scale all three values by the same factor.

Consequently, the determinant must be zero, and we get the condition

\begin{eqnarray} D &~=~& a_2\cdot a_3\cdot w_1\cdot w_3\cdot c_1 + a_1\cdot a_2\cdot w_2\cdot w_3\cdot c_3 + a_1\cdot a_3\cdot w_1\cdot w_2\cdot c_2 \\ && {} - a_1\cdot a_2\cdot w_1\cdot w_3\cdot c_3 - a_1\cdot a_3\cdot w_2\cdot w_3\cdot c_2 - a_2\cdot a_3\cdot w_1\cdot w_2\cdot c_1 ~=~ 0 \end{eqnarray}

As this equation is linear in each of the nine values $c_1,c_2,c_3$ and $a_1,a_2,a_3$ and $w_1,w_2,w_3$, it easily allows to compute any of these nine values from the other eight values.

• Originally misread "$\text{COW_GRASS_PER_WEEK}=\text{Quantity of grass eaten by a cow in one week}$" as "Quantity of cows eaten by grass in one week". Which, for me, answered the question of "Best / worst idea for revenge horror movie" ("The grass... It's fighting back!"). Anyway, upvote partly because of that. – Parthian Shot Mar 4 '16 at 18:39
• If the determinant was nonzero, the matrix would be invertible and you would have a unique solution for CGW, AGW, and IGA. But you can scale all three by the same factor, so the solution isn't unique. The determinant must be zero. – f'' Mar 4 '16 at 18:53
• Thanks to everyone who edited and completed the answer, I left it as a WIP but had no time to finish it... Usually I can spend some of my work time on PSE, but in this days I had to actually work (hate when it happens) – Narmer Mar 9 '16 at 14:20

We have $a_i$ acres of land, $c_i$ cows and $w_i$ weeks. Since the grass grows uniformly over the field, the $a_i$ also measures the amount of grass in, lets say, acre-equivalent mass.

Now lets denote with $f$ the cow feed rate in acre-equivalent/week and with $g$ the grass growth rate in acre-equivalent/week.

We want the cows to eat all the grass just at the end of our time frame. So we can write: $$a_i + a_i w_i g - c_i w_i f = 0$$ The first term is the amount of grass available at the beginning, the second term describes the growth of grass and the third term the amount of grass the cows eat.

As per the problem, we have three sets of quantities and so three equations: $$a_1 + a_1 w_1 g - c_1 w_1 f = 0$$ $$a_2 + a_2 w_2 g - c_2 w_2 f = 0$$ $$a_3 + a_3 w_3 g - c_3 w_3 f = 0$$ Lets assume that the first two sets are given, like in the example above.

From the first equation we can now express the grass growth rate: $$g = \frac{c_1 w_1 f - a_1}{a_1 w_1}$$ This can now be inserted into the second equation, from where we can get the cow feed rate: $$f = \frac{a_2 - a_1 \frac{a_2 w_2}{a_1 w_1}}{c_2 w_2 - c_1 w_1 \frac{a_2 w_2}{a_1 w_1}}$$ With this, both of the parameters are fixed.

Depending on the quantity missing from the third set, we have to use one of the below equations: $$a_3 = \frac{c_3 w_3 f}{1 + w_3 g}$$ $$c_3 = \frac{a_3 + a_3 w_3 g}{w_3 f}$$ $$w_3 = \frac{a_3}{c_3 f - a_3 g}$$