Puzzles

When halfway around the corner, the grand piano will look like this:

The problem then is finding the largest area rectangle which fits in the highlighted triangle: an isosceles triangle where the base is twice the height. Let the base be 2 and the height be 1.

If the height of the rectangle is \(h\), then its width is \(2(1-h)\). Therefore its area is \(2h-2h^2\). By differentiation, it can be seen that this is maximum when \(h=\frac{1}{2}\), which means that ratio of the rectangle's length to its width is 2:1.

If the corner was not a 90° angle, then what is the largest area rectangle which could fit round it?