来源: 经济数学 Alpha C. Chian 科目: 经济数学　Ch12 Optmization with equality Constrains 问题: 题目中说的12.22式 f(v)≧f(u) => { Σfj(u)(vj-uj) } { Σfj(v)(vj-uj) } ≧0 { } The function z=f(X1,X2)= X1X2 (X1,X2 ≧0） is quasiconcave . (半凹的吗?) We shall now check it. Let u= (u1,u2) and v=(v1,v2) be any two points in the domain. Then f(u)=u1u2 and f(v)=V1V2 assume that: f(v)≧f(u) or v1v2≧u1u2 (v1 v2 u1 u2≧0) (12.27) since the partial derivitives of are f1=x2 and f2=x1 12.22式 amounts to the condition that (请问这边f的偏导数根下面推出的方程式有何关系?) f1(u)(v1-u1)+f2(u)(v2-u2) = u2(v1-u1)-u1(v2-u2) ≧0 (为何透过导数可以判定需要都以 u去测度? 以及f1(u)和f2(u)为何等於 u2和u1 ?? 下面12.28的式子怎麽来的) or upon rearangement , u2(v1-u1)≧u1(u2-v2) (12.28) we need to consider four possiblities regarding the value of u1 and u2 First, if u1=u2=0 then 12.28 is trivially satisfied. Second , if u1=0 but u2>0 then 12.28 reduce to the condition u2v1≧0. which is again (违反的意思吗?? 那为何可推出使u2 v1不为负成立) satisfied since u2 and v1 are both nonnegtive. Third , if u1>0 and u2=0 then (12.28) reduce to the condition 0≧-u1v2, (也同样是违反0≧-u1v2吗 那又为何等是成立?) which is still satisfied. Last, suppose that u1 and u2 are both possitive so that u1 and u2 are both possitive , so that v1 and v2 are also positive. (这边比较没问题) Subtracting v2u1 from both sides of (12.27), we obtain v2(v1-u1)≧u1(u2-v2) (12.29) (这边还看的懂 只是不知道同时减掉 v2u1的用意??) Three subpossibilities now present themselves. (不懂这个单字?? 推测??) 1.If u2=v2 then v1≧u1 .In fact , we should have v1>u1 since (v1,v2) are distinct points. (不懂?? u2=v2不是会使等号右边变成0? 那怎麽又推得 v1≧u1 以及distinct point 的 v1>u1? 不是也应该是 v1=u1?? The fact that u2=v2 and v1>u1 implies that condition(12.28 is satisfied) 2.If u2>v2 , then we must also have v1>u1 by(12.29) .Multiplying both sides of (12.29) by u2/v2 , we get u2(v1-u1)≧ u2/v2[u1(u2-v2)]>u1(u2-v2) [since u2/v2 >1 ] (12.30) Thus 12.28 is again satisfied (这边我看的懂 我比较没意见) 3. The final subpossiblities is that u2<v2, implying that u2/v2 is a positive fraction. In this case . the first line of (12.30) still holds. The second line also holds , but now for a different reason: (以上我还能够理解 下面两句就看不懂 ) a fraction (u2/v2) of a negative number (u2-v2) is greater than the latter number itself. 请问 整个题目的中心主旨 就是在映证 原本 透过12.27的假设而来的 12.28恒等式吗 u2(v1-u1)≧u1(u2-v2) 题目太复杂了 又有用到一些特别的英文 烦请高手 感激不尽 -- Anything, without blind insistence, is right. --

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