Maxzimize the function 最大化函數問題

a, b=0,1,2,...,N-1, (a,b)≠(0,0), find max. of f(a,b)=a/(a+b)- a/(2N-a-b)
1. if a+b>=N, then f(a, b)<=0, so we can suppose that a+b<=N
2. ∂f/∂b= -a/(a+b)^2 - a/(2N-a-b)^2 <= 0
so, f is increasing w.r.t. b (any a>=0)
finding max. of f(a, b), so b=0, f(a, b)=g(a)=1-a/(2N-a)
3. g'(a)= -2N/(2N-a)^2 <0, so g(1) is max. (Note: (a,b)≠(0,0))
Ans: max. f(a,b)=f(1,0)=1-1/(2N-1)= 2(N-1)/(2N-1)



2010-02-27 22:28:54 補充：
(Suppose a+b < N, then a+b < 2N-a-b and a+b+1 <= 2N-a-b-1)
f(a,b)-f(a,b+1)=a/(a+b)-a/(2N-a-b)-a/(a+b+1)+a/(2N-a-b-1)
=a/[(a+b)(a+b+1)] - a/[(2N-a-b)(2N-a-b-1)] >= 0
so g(a)=f(a, 0) > f(a, b)
g(a)-g(a+1)= 2aN/[(2N-a)(2N-a-1)] > 0 for a=1,2,3,...N
so max. of f(a,b)=f(1,0)

try not to use calculus