Unformatted text preview: → West distance when the barrier is crossed, and the mean increase given that one occurred would be less than 2L/3. If the barrier is x units north of the city’s Southern border, then the city is divided into two rectangles, one of area Lx and the other of area L(L-x). (We assume, of course, that x < L.) Thus, if two points are chosen at random over the city, the probability that one of them is north of the barrier and the other south is 2Lx[L(L-x)] /L 4 . (See why?) This quantity is a numerical constant times x(L-x). The value of x that maximizes the expression is found by setting the appropriate derivative equal to 0, which quickly leads to the outcome x = L/2. In short, the barrier should divide the city into two halves, and travelers between the two sectors should have to travel to the edge of the city to get through. As in Berlin, however, such a barrier cannot be expected to last forever....
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This note was uploaded on 11/08/2011 for the course AERO 16.72 taught by Professor Hansman during the Fall '06 term at MIT.
- Fall '06