14 Mar 07
Originally posted by Shallow BlueWell actually the problem stated that the fence had to be straight apart from sharp corners. If it didn't then obviously the best solution is a circle (A=2500/pi) [as my first problem related to an open field with no river].
Does it have to be a polygon? I suspect, but cannot prove, that a semi-circular or semi-elliptical fence would be optimal.
Richard
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14 Mar 07
Originally posted by XanthosNZThen you'll have to define "sharp corners". It can't mean an acute angle, since the previous solutions used right angles (and anyway, if you allow only acute angles the solution trivially becomes "as closely approaching the right angle solution as you can manage with this fencing"😉. IYAM if you allow acute and right angles but no obtuse ones the solution must be the rectangular one; and if you do allow any obtuse angle, you start the approaching trick again, approaching a circle as closely as possible.
Well actually the problem stated that the fence had to be straight apart from sharp corners. If it didn't then obviously the best solution is a circle (A=2500/pi) [as my first problem related to an open field with no river].
Richard
14 Mar 07
Originally posted by Shallow BlueJesus, stop reading into it. A square has sharp corners, so does an octogon and so does a regular polygon with 100 sides. A circle does not.
Then you'll have to define "sharp corners". It can't mean an acute angle, since the previous solutions used right angles (and anyway, if you allow only acute angles the solution trivially becomes "as closely approaching the right angle solution as you can manage with this fencing"😉. IYAM if you allow acute and right angles but no obtuse ones the solutio ...[text shortened]... start the approaching trick again, approaching a circle as closely as possible.
Richard
Note: Without the allowance of 0.5m the answer is trivial (as many sides as you can have as you start to approximate a solution), it becomes an optimization problem because you must balance the approximation of a circle with a reduced fence length with more corners.
14 Mar 07
Originally posted by mtthwThanks mtthw! Just tried a little Lagrange multiplier calculation here...almost ran out of ink. 😕 I think it might be easier to show that for a given perimeter and a given number of sides, say 3, that an equilateral triangle covers more area than any other type of triangle, and then proceed with a proof by induction.
I think you're looking for Lagrange multipliers.
To find the maximum of f([b]x), subject to the constraint g(x) = c,
Maximise F(x, lambda) = f(x) - lambda*[g(x) - c]
But I'm not planning on trying the calculation myself![/b]
20 Mar 07
Originally posted by PBE6Not getting anywhere right now...XanthosNZ, got a hint for me?
Thanks mtthw! Just tried a little Lagrange multiplier calculation here...almost ran out of ink. 😕 I think it might be easier to show that for a given perimeter and a given number of sides, say 3, that an equilateral triangle covers more area than any other type of triangle, and then proceed with a proof by induction.
21 Mar 07
Originally posted by PBE6If for a given number of vertices and fixed total edge length the polygon that encloses most area is the regular one then since you've already shown that the nonagon is the maximum regular case it's proven.
Not getting anywhere right now...XanthosNZ, got a hint for me?
21 Mar 07
Originally posted by DeepThoughtNo, the second part...proving that a regular polygon constructed according to the stipulations encompasses the greatest area, greater than any irregular polygon. I'm still working on it...hopefully I'll come up with something before my teeth fall out. 😕
If for a given number of vertices and fixed total edge length the polygon that encloses most area is the regular one then since you've already shown that the nonagon is the maximum regular case it's proven.
21 Mar 07
Originally posted by PBE6You've shown that the regular 9-gon is the optimal regular polygon. What remains, I believe, is to show that for any n-gon, for a set perimeter length, the regular n-gon maximizes the area.
No, the second part...proving that a regular polygon constructed according to the stipulations encompasses the greatest area, greater than any irregular polygon. I'm still working on it...hopefully I'll come up with something before my teeth fall out. 😕
I didn't have any clue how to prove this so I did some google searching and I believe I found enough pieces to show this. Since I've "cheated" I won't spoil the problem--but I believe there is a fairly nice proof. PM me if you want the spoiler.
21 Mar 07
4 edits
Originally posted by PBE6I probably didn't express this very well. What I was getting at is that you don't have to worry about the stipulations. Once you've fixed the number of vertices you've fixed the perimeter length. The polygon with the largest area for a given perimeter length is the regular one. Since you have already shown which regular polygon is the one with largest area we are home and dry; unless you want to prove that the regular polygon is the one with largest area.
No, the second part...proving that a regular polygon constructed according to the stipulations encompasses the greatest area, greater than any irregular polygon. I'm still working on it...hopefully I'll come up with something before my teeth fall out. 😕
I can't prove it either, except that for n-gons with the same perimeter the n-gon with maximum area must be convex; any concave bits can just be reflected so that they're convex and the area increases each time you do that. Are we allowed to assume that a circle has the largest area of any shape with a given perimeter length - if so you can show it's cyclic as well.
21 Mar 07
Originally posted by DeepThoughtYou're not reading very well, either. That's exactly what I'm trying to do! Geez, just read leisurelysloth's post if you refuse to understand mine.
I probably didn't express this very well...Since you have already shown which regular polygon is the one with largest area we are home and dry; unless you want to prove that the regular polygon is the one with largest area.
21 Mar 07
Without the allowance of 0.5m the answer is trivial (as many sides as you can have as you start to approximate a solution)What's all the fuss about here? As Xanthos has conceded "Without the allowance of 0.5m the answer is trivial (as many sides as you can have as you start to approximate a solution)". So with or without that stipulation, the closest approximation to a circle is a regular convex polygon with as many sides as possible.