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
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° * | * | * | * * |