# Math Help - Quickie #11

1. ## 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:
      +----*-----
|     *
|      *
|       *
|        *
|     *
|  *
*
|

2. 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:

$A=\frac{x \cdot y}{2}+\frac{1}{2} \cdot \sqrt{x^2+y^2}\cdot \sqrt{4^2 - \left( \frac{1}{2}\sqrt{x^2+y^2} \right)^2}$

$A=\frac{x \cdot y}{2}+\frac{1}{2} \cdot \sqrt{16 (x^2+y^2) - \frac{1}{4}(x^2+y^2)^2}$

$A=\frac{x \cdot y}{2}+\frac{1}{4} \cdot \sqrt{(x^2+y^2)\cdot (64-(x^2+y^2))}$

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

EB

3. 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??

4. 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°  *
|           *
|       *
|   *
*
|