Thread: Quickie #11

Across one corner of a rectangular room, two 4-foot screens
are placed so as to enclose the maximum floor space.
Determine their positions.

Code:

      +----*-----
      |     *
      |      *
      |       *
      |        *
      |     *
      |  *
      *
      |

Originally Posted by Soroban

Across one corner of a rectangular room, two 4-foot screens
are placed so as to enclose the maximum floor space.
Determine their positions.

Code:

      +----*-----
      |     *
      |      *
      |       *
      |        *
      |     *
      |  *
      *
      |

Hello Soroban,

unfortunately I can't offer you a complete solution..

At first sight I would say that there must be 2 different solution to place those screens:
a) they form with the walls a square with area = 16 sqft
b) they form a straight line which is the base of an isosceles right triangle with area = 16 sqft

At the attempt to get a more general solution I got lost completely:

According to the attached sketch I got:







At the attempt to substitute y by a term in x I failed successfully. I only can ask you to ban my sorrow

EB

Maybe a trick question?

Each are 4 feet away from the corner to make a square of 16 square feet?

Hehe...


Edit: Or it could be infinity because when the area is enclosed, the area enclosed by the screens are on the outside??

Nice try, EB . . . and a good guess, anthmoo!

Remember this is a "Quickie",
. . so there is usually a surprisingly simple solution.
I found all of them to be valuable learning experiences.

This problem has a head-slapping, OMG kind of solution . . . brace yourself.


We have a pair of perpendicular walls and a pair of 4-foot screens.
They could be arranged like this:

Code:

      * - - -*- - - - - -
      |       *
      |        *
      |         *
      *          *
      |   *        *
      |       *     *
      |           *  *
      |               *

Now consider four of these figures clustered about the right angle.

Code:

      *       |       *
       *  *   |   *  *
        *     *     *
         *    |    *
      - - * - + - * - - 
         *    |    *
        *     *     *
       *  *   |   *  *
      *       |       *

We have an octagon with eight equal sides.

Obviously an octagon with a fixed perimeter has maximum area
. . when it is a regular octagon.


Therefore, the screens are placed like this:

Code:

      *-----------------*--
      |                *
      |               *
      |              *
      |             *
      |      135°  *
      |           *
      |       *
      |   *
      *
      |