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:+----*-----
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Printable View
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:+----*-----
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Hello Soroban,Quote:
Originally Posted by 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 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:* - - -*- - - - - -
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Now consider four of these figures clustered about the right angle.Code:* | *
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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:*-----------------*--
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