# Hinged Screen

Bert Bates

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Sheriff

Sheriff

Posts: 8919

11

posted 13 years ago

A hinged screen with two panels of the same size is placed into the (90 degree) corner of a room. At what angle should the panels be arranged so that the corner and the screen enclose the greatest area?

[ November 16, 2003: Message edited by: Bert Bates ]

[ November 16, 2003: Message edited by: Bert Bates ]

Spot false dilemmas now, ask me how!

(If you're not on the edge, you're taking up too much room.)

Jim Yingst

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Sheriff

Sheriff

Posts: 18671

posted 13 years ago

Hmm.

If the width of one panel is x, and the angle of the hinge is 90 degrees, then the area enclosed will be x^2, for you will have formed a square of side length x.

If you make the angle of this hinges 180 degrees (the two panels form a straight line), then you can make a 45-45-90 right triangle in the corner with a base of 2x and legs x * sqrt(2), creating a triangle with base 2x and height x, and thus having an area of x^2 (same as the square).

However, if you have an angle of 135 degrees, then you have created a shape that can be described as two trianlges that share the same base. That base is of length approx. 1.86 * x, and the two trangles have areas of .35 x ^2 and .85 x ^2, summing for an area of 1.2 x ^2

Since both extremes have been demonstrated to have the same value, (and since I never did do much with calculus), I will take as the maximum possible the midpoint between the two extremes (135 degrees), with an area of 1.2x^2.

If the width of one panel is x, and the angle of the hinge is 90 degrees, then the area enclosed will be x^2, for you will have formed a square of side length x.

If you make the angle of this hinges 180 degrees (the two panels form a straight line), then you can make a 45-45-90 right triangle in the corner with a base of 2x and legs x * sqrt(2), creating a triangle with base 2x and height x, and thus having an area of x^2 (same as the square).

However, if you have an angle of 135 degrees, then you have created a shape that can be described as two trianlges that share the same base. That base is of length approx. 1.86 * x, and the two trangles have areas of .35 x ^2 and .85 x ^2, summing for an area of 1.2 x ^2

Since both extremes have been demonstrated to have the same value, (and since I never did do much with calculus), I will take as the maximum possible the midpoint between the two extremes (135 degrees), with an area of 1.2x^2.

Piscis Babelis est parvus, flavus, et hiridicus, et est probabiliter insolitissima raritas in toto mundo.