The Pirates' Cantilever

Solution to Last Month's Dr. Ecco

Solution to "Optimal Farming," DDJ, May 2005.

1. 1. Observe that the minimum circumscribing circle for a square having side 2L has radius L√2 and area ×2×L2, so the extra area is (-2)×2×L2. This is divided equally among the four sides. So each side gets (-2)×(L2)/2 extra area. In the design of Figure 1, the small squares have side lengths 2L where L=1/4. The circles of radius L have 10 extra sides having total area of 5×(-2)×(1/16). The large square has L=1/2, so has (-2)/2 extra area. The total cost is slightly less than (13/16)×(-2).
2. 2. For the case where the circles all must have the same radius, try the design where there are four squares each having side length 2L. There are two illustrated in Figure 2. Then there are circles covering each of these squares. So the radius of the circumscribing square is L√2. (Note that the middle of the rectangle is 1-2L from each side square.) In addition, there is a circle having a radius L√2 whose center is the middle of the rectangle. That reaches the near-center corner of the corner square because that distance is (by the Pythagorean theorem) √((1-2L)²+(1/2)²)=√(1-4L+4L2+1/4)= √(5/4+4L2-4L). The length is L×√2. Squaring that gives 2L2. So 2L2=4L2- 4L+5/4, yielding the quadratic expression 2L2-4L+5/4=0. So, L= 4/4+/-√(16-10)/4=1-(√6)/4.

M.S. Gopal contributed greatly to these solutions.