When I read this, the first thing that popped into my mind was "I thought the rule of thumb was that for a given perimeter, a circle encloses the maximal area". This led me down a path of thought which lasted throughout morning shower, and has left me with three things: (1) a realization that my arithmetical apparatus, sans calculator, has rusted to the point of dereliction (squaring things is _hard_! and dividing by pi? forget about it), (2) something of an answer, and (3) some very wrinkled skin.
I didn't get as far as a general answer which accounts for all barn sizes. I started off by calculating the ideal barn length to allow 100m of fence to describe a perfect semicircle, and assumed that barn size thereafter. I'm not sure if this biased my results by giving an advantage to the circle, but if it is fair, the results support the original premise (that the semicircle is optimal). Also, if we think in terms of feet instead of meters, it actually gives a reasonable barn size. Here's how I progressed. First, let's calculate the area enclosed by a perfect semicircle whose curved perimeter is 100. Ok, the area of a circle is 1p1 * r^2, so we need r . The radius can be derived from the perimeter with the relation p = 2 * 1p1 * r . We're only dealing with a semicircle, so remove a factor of two (hey, here's a situation where pi makes calculation easier than tau!). So we have 100 = 1p1 * r and r = 100%1p1, about 32 (and since the barn wall is the circle's diameter, the barn is about 64 long, which sounds reasonable - in feet). So now we can start calculating and comparing the areas of different geometries. The semicircle's area is half of the corresponding circle's, so 1r2*1p1*32^2 or ~1600 . To compare that to a quadrilateral, let's run fence parallel to the length of the barn, and then connect it with two identical, shorter stretches, for the sides. The long side is 64 (to match the barn) and each short side is 1r2 * (100-64) or 18. So the total area is 32*18, which is < 1200, and loses to the circle. Ok, but there are many ways to make a quadrilateral, and maybe some have larger areas. But we made the only possible rectangle, so the only options left (other than just skewing the rectangle, which won't change its area) involve angling the sides inwards and shortening the fence opposite the barn. So let's just take that to the extreme and try an isoceles triangle. We need two stretches of equal length, so each side of the triangle will be 50, and its base will be 64. Its area is 1r2 * base * height. To calculate the height, we bisect the triangle into two right triangles, and apply the Pythagorean theorem, where the hypotenuse is 50, the base is half the barn's length (32), and height = %: (50^2)-(32^2), which comes out to about 39 . So the area enclosed by our isoceles triangle is about 32*39 or ~1250 . So the semicircle still wins. What else can we do? What other geometries could credibly improve over the semicircle? A pentagon? Hexagon? Those might improve over the triangle and rectangle, but they just seem like incremental advancements towards the semicircle. And asymmetric designs, or geometries with interior partitions, or other esoterica, just don't seem credible ways to increase area (at an intuitive level, anyway). Plus the semicircle has practical advantages: it won't gather gunk built up of detritus blowing around the farmyard, and it will be easy to sweep out and keep clean. Though it's going to be harder to corner a chicken. Anyway, that's where I stopped - it was really time to get out of the shower, I was becoming a prune. -Dan Please excuse typos; composed on a handheld device. On Feb 23, 2013, at 9:42 AM, km <k...@math.uh.edu> wrote: > Use J to solve the farmer's fence problem: > > A farmer with 100 meters of wire fence wants to make a rectangular chicken > yard using an existing barn wall for one of the north-south sides. What is > the largest area he can enclose if he uses the 100 meters of fence for the > other three sides, and what are the dimensions of the largest-area chicken > yard? > > Kip Murray > > Sent from my iPad > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
