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There is an earlier question, The grazing cows of Sir Isaac Newton, which says:

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.

I would like to ask a further question about dynamic equilibrium. Given this setup, what is the minimum area required to sustain 10 cows for infinite time?

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    $\begingroup$ Consider 1 cow : "how much it eats in a week" must be equal to "how much grass grows in A acres in a week" , which shows that area 10*A is required for 10 cows. Eaten grass gets replaced by growing grass hence it will go on for "infinite time" $\endgroup$ – Prem May 10 '18 at 12:35
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    $\begingroup$ Provided sufficient grass, do the cows live forever, or do we need to account for a bull, breeding strategies, and predators? $\endgroup$ – Ian MacDonald May 10 '18 at 13:17
  • $\begingroup$ @IanMacDonald , nice catch ! I too had considered it and hence I used quote infinite time unquote ! If we were to go that way , then we need to consider which solar-system/universe it happens in , because some theories say that our sun will be gone in a few billion years and later , our universe itself will go out. I think short term steady state should be enough for this puzzle. $\endgroup$ – Prem May 10 '18 at 17:37
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    $\begingroup$ Welcome to Puzzling! (Take the Tour!) I'm glad you're eager to contribute here, but this question as it stands is heading quickly toward closure for being incomplete (and thus "unclear what you're asking"). Questions should really be self-contained; too much of this current question relies on the original question link to stand on its own. I've added the relevant setup from the original question to this one, so it should now be clear what is being asked. (Whether it's now just a math problem vs. a puzzle I'll leave to others to decide.) $\endgroup$ – Rubio May 10 '18 at 19:50
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According to the solution of the other question, the CGW = $\frac{10}9$ AGW, meaning the grass eaten per cow per week is $\frac{10}9$ times the amount of grass that grows per acre per week.

Thus, 10 acres can support 9 cows indefinitely, and it will take $11 \frac19$ acres to support 10 cows indefinitely.

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