|
Max=2^N. Min=1+N.
As Federico Kereki pointed out to me as this problem was being evaluated in the queue, the proof is essentially just the inequality (1+x)*(1+y)=1+x+y+x*y > (1+(x+y)).
Productivity can always be improved by staffing another subproject as long as that is possible. For if we suppose that there are less than N subprojects, then at least one of them is staffed by x > 1 engineers, and taking an engineer from such to staff another subproject will replace (1+x) with the larger value (1+(x-1))*(1+1).
Similarly, merging two subprojects will always lower the productivity, since with x engineers on the one and y on the other, (1+x)*(1+y) > (1+(x+y)).
The highest productivity thus results when there are N subprojects (each staffed by a single engineer), and the lowest when the project has not been split up, since any assertion that an extreme value of productivity occurs for k subprojects, 1 < k < N, can be immediately refuted.
Notice also that (1+x_1)*(1+x_2)*...*(1+x_k) = 1+x_1+x_2+...+x_k+Y = (1+N)+Y where Y>=0, and Y=0 only if k=1, giving another proof for the minimum. |
