That Stingy Rancher And His Fence Again
A rancher has a square mile of flat land, a perfect square with a fenced perimeter of 4 miles. He decides he wants to divide his property into 4 parts of equal areas. He knows he can do this with 2 miles of fencing, which is $1000 per mile, but he is feeling particularly stingy and thinks he can do it with less. Let A = the total length of fence. A is minimized. How much did he pay for fencing, expressed as
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1
0
0
0
A
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This problem can be done without calculus. The 4 parts need not be of the same shape. The answer is less than 2 0 0 0 .
The answer is 1975.
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1 solution
Speechless... How did you think of this? What was the thought process?
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Calvin, I was just about to put that down when i got a phone call. This one is an oldie from way back, in my files. Here are the salient points that help arrive at this solution:
1) Any segment is always an arc of a circle (which includes straight lines)
2) Any segment ending at a fixed line (straight or curved) is always perpendicular to it
3) Any vertex can only have at most 3 segments coming together there, and they must all be at 120 degrees apart
Only a few possible configurations are possible meeting these conditions (after considering symmetry), and it's just a matter of finding which yields the least fencing. Jakob Steiner wrote extensively on such minimization problems back in the 19th century, and those were his findings.
How to think about such problems? Think of how a soap bubble would do it. One of the most fascinating things about soap bubbles is how instantly they can achieve minimal configurations that can take a computer a long time to figure out. I think a lot of physics is lilke that.
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I'm aware of those conditions, and of soap bubble physics.
The amazing part is that nature does these "complicated equations" instantaneously, by seeking equilibrium / maximization.
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@Calvin Lin – Much of physics is about symmetry and minimization/maximzation. This kind of stuff appears almost anywhere you look in theoretical physics. I do think the reasons why we have certain laws, such as, "Princple of Least Action", are due to processes similar to what soap bubbles do.
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@Michael Mendrin – Dear Michael I guessed the shape but I am not able to understand how to find these values can u please explain in detail...
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@Hemant Khatri
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Imgur
Hemant, did you get that for the arc to start at 120° to the 45° diagonal line and meet the side at right angles, it has to be 15°? Here is how one might proceed -
Areas of sector JCB; Triangle JCB and the 'cap' = 2 4 π R 2 ; 2 R 2 sin 1 5 ; R 2 ( 2 4 π − 2 sin 1 5 ) Area of triangle VJB = sin 4 5 2 R 2 sin 5 2 . 5 × ( sin 7 . 5 ) 2 × sin 8 2 . 5 Equate the sum of areas of triangle VJB and the Cap to one eighth to get the arc radius: R = 2 4 π − 2 sin 1 5 + sin 4 5 2 sin 5 2 . 5 × ( sin 7 . 5 ) 2 × sin 8 2 . 5 0 . 1 2 5
Just curious, but where are all the arc centers located?
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Well, since the arcs have to end perpendicularly with the sides, all the centers of the arcs are on [extended lines of] the sides. One way to solve this problem: Let x be the distance one of those centers is from some origin on the same line. Find out where A is. Find area of ABCD. Solve for x such that ABCD = 1/4. Etc. There's lots of ways to go about this.
Amazing! I checked this out with an Excel Spreadsheet and it verifies the concept although I got 1976. Picky, picky!
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I think what's even more astonishing is that you were able to do this on Excel. But thanks for the confirmation.
However, I checked my notes, the answer does round off to 1976. But convention here in Brilliiant is to go with floor values when asking for integer answers.
Well I got 1976 by rounding it but thanks to your hint of Floor function in questions. got it right at very beginning. Amazing thought. This was an interesting problem.
https://brilliant.org/problems/find-the-areaonly-your-logic-can-help-you/?group=3UHxOzwinQpA&ref_id=384997
PLEASE TRY TO DO THIS AWSOME PROBLEM TOO..post a solution if you get......................i am waiting for an awesome solution that i made while creating this problem
Here's the graphic of how this is done
Determine where point A should be on the diagonal such that the area of parcel ABCD is exactly one quarter. Then work out total length of fence.