|
|
Re: variation on a familiar calculus problem
Posted:
May 2, 2012 9:30 PM
|
|
Sandy, I think I have come up with an argument that actually applies to both of your cases (the rectangular fence against the wall and the semi circular fence against the wall).
For the case of the rectangular fence: We are asked to construct 3 sides of a rectangle against a wall using a fixed amount of fencing such that the enclosed area is the maximum area possible. Let's assume that we have constructed this rectangle and then we construct the same rectangle on the opposite of the wall (a mirror image) such that we have one large rectangle (ignoring the wall). Since each of the smaller rectangles are at maximum area (our assumption) then the large rectangle must also be at maximum area, which means that the large rectangle is a square (because a square attains the maximum area for a fixed rectangular perimeter). Therefore, the rectangle on one side of the wall must be half a square.
The same reasoning applies to the semi circular fence with the assumption that we know that a full circle attains the maximum area for a fixed (rectangular or curved) perimeter. Thus, whatever we create on one side of the wall can be mirrored on the other side of the wall. Since the shape on one side of the wall is a maximum then so should the combination of that shape with its mirror image on the other side of the wall. The full shape must then be a circle and the shape on each side of the wall, a semi circle.
This all relies on knowing that a square is the best rectangular shape to enclose the most area and that a circle is the best of any shape to enclose the most area. Your two problems (with a wall) are essentially to find "half" of the best of those two cases.
Bob Hansen
On May 1, 2012, at 8:36 PM, Sandy Wagner <papasandy@comcast.net> wrote:
> (I'm a retired HS math teacher now working with one student at a time)
>
> A student and I explored the familiar question of maximum area with L units of fence with one side of the enclosure being a straight wall. The rectangle with max area would have the ratio of sides 2:1. Then I asked the same question with the allowed shape being a portion of a circle (see attached). I had always wondered about this problem but never got around to it. The solution surprised me because of its simplicity - it's a semicircle with radius L/pi.
>
> My question is this: could that solution have been guessed? That is, is there an informal analysis that would lead to the solution?
>
> Thanks
> -Sandy Wagner
> Menlo Park, CA
> <circle problem.jpg>
|
|
