your assessment of the semicircle winning is correct (1591m^2).
Your assessment of the isosceles triangle assumes that the base will
be 64m- but that is not known and can't be assumed- what you do get is
the equivalent of a rectangle which is 32 by 38.5-which unfortunately is
>100m of wire.
Given the requirement of a rectangular enclosure we can't assume this
or any general quadrilateral- we must have a rectangle.
we know that for a rectangle with side on the barn =b (and that opposite
the barn as well) and the sides perpendicular to the barn of length x
we have two expressions
b+2x=100 or b=100-2x
and
A(rea)=b*x
so A=(100*x) -2**:x
differentiate and get dA/dt=100-4x =0 for the maximum (second derivative
is -ve so A is a maximum)
this gives x=25m so as a result b must be 50m (note that this implies
that the barn is at least 50m long- the problem doesn't stipulate that
the whole side of the barn must be used).
and the area A= 1250m^2
Alternatively choose a set of values x, find b and A (use 0<x<+100r3)
and plot A vs x
x,. b,. a=:b*x
5 90 450
10 80 800
15 70 1050
20 60 1200
25 50 1250 <------
30 40 1200
32 36 1152
Now, I do think that you can express this in a J sentence better than I
can-which I believe is the object of the exercise
Don Kelly
On 23/02/2013 9:01 AM, Dan Bron wrote:
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
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