# Quickie #11

• Jan 12th 2007, 07:36 AM
Soroban
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:

      +----*-----       |    *       |      *       |      *       |        *       |    *       |  *       *       |
• Jan 14th 2007, 12:03 AM
earboth
Quote:

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:

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

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

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

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

EB
• Jan 14th 2007, 01:26 AM
anthmoo
Maybe a trick question? :rolleyes:

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??
• Jan 14th 2007, 02:35 AM
Soroban
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°  *       |          *       |      *       |  *       *       |