What is the maximum fractional coverage of the area of a circle by three, non-overlapping rectangles, all of which lie internal to the circle? (The rectangles can be of different sizes)
Lay the rectangles horizontally and symmetrically, corners touching
the circle, with one large central rectangle sharing the center
of the circle, (0,0), and two smaller identical rectangles above and
below. (I have no proof why
this should be optimal).
Maximize 1/2 area large + area small in the top semicircle.
Say the large rectangle has half height b. (We will get the
area sum in terms of b alone)
The area of large rect, height * width = 2b * 2(1-b^2)^1/2.
1/2 A_Large = 2 b (1-b^2)^1/2
Top right corner of top small rectangle is (x,y) on the circle, so
its area is:
A_small = 2x * (y-b) = 2 * (1-y^2)^1/2 * (y-b)
Imagine b as a constant. Maximize the area of the small rectangle,
expressing y in terms of b. Differentiate wrt y:
2 (1-y^2)^1/2 - (1-y^2)^(-1/2)(-2 y) (y-b) = 0
2 (1-y^2) + 2by - 2y^2 =0
2 y^2 -by -1 = 0
choose positive root
y = [b + (b^2 + 8)^1/2]/4
So, the answer is gotten by maximizing:
A_tot = 1/2 A_Large + A_small over b, where:
A_tot = 2 b (1-b^2)^1/2 +
2 * (1 - {[b + (b^2 + 8)^1/2]/4}^2) * ({b+[b^2 + 8]^1/2}/4 -b)
Not difficult, just lengthy.
Edited on April 29, 2019, 12:28 pm
most of the way....
