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

Printable View
Across one corner of a rectangular room, two 4foot 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? :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??
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 headslapping, OMG kind of solution . . . brace yourself.
We have a pair of perpendicular walls and a pair of 4foot 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° *
 *
 *
 *
*
