Dido's Problem/Variant 2

Problem

Consider the corner of a room which is separated from the rest of the room by a screen of two identical halves, $a$ and $b$ where the length of $a$ equals the length of $b$.

The screen meets the walls of the room at $A$ and $B$.

How can the screen be arranged so that the area enclosed by the screen is a maximum?

Solution

When the screen forms two sides of a regular octagon:

Thus the sides of the screen must make an angle of $67 \frac 1 2 \degrees$ with the walls.

Proof

When the area enclosed by the screen is a maximum, so will be the area of the octagon so formed by the construction above.

The angles the screen makes with the walls is half the internal angle of the regular octagon.

From Internal Angles of Regular Polygon, that is:

$A = \dfrac 1 2 \paren {\dfrac {\paren {n - 2} 180 \degrees} n}$

where:

$A$ is the angle the screen makes with the wall
$n$ is the number of sides of the regular polygon, in this case $8$.

Hence the result.

$\blacksquare$