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Re: Finding the optimum by repeteadly zooming on the solution space (or something like that)
*To*: mathgroup at smc.vnet.net
*Subject*: [mg92575] Re: Finding the optimum by repeteadly zooming on the solution space (or something like that)
*From*: Bill Rowe <readnews at sbcglobal.net>
*Date*: Mon, 6 Oct 2008 04:14:14 -0400 (EDT)
On 10/5/08 at 6:05 AM, cuak2000 at gmail.com (Mauricio Esteban Cuak)
wrote:
>Hello everyone.I know somebody has probably writ mathematica code on
>this problem. A general answer would probably benefit more people,
>but I'll do the specific example, 'cause I'm not sure how to explain
>it in another way.
>I have these restrictions:
>cpoAg1= ( -x + (70*a*(x^0.4 + y^0.4)^0.75)/x^0.6);
>(*and*)
>cpoAg2= ((70*(1 - a)*(x^0.4 + y^0.4)^0.75)/y^0.6 - 2*y);
>(* And I need to find the "a" that will maximise this following
>function. I don't need and exact solution, but something that is
>sufficiently near *)
>obj = -x^2/2 + 100*(x^0.4 + y^0.4)^1.75 - y^2;
>(* "A" goes between 0 and 1 so I discretise to obtain the {x,y} that
>maximise every "a"
>the Table command gives me the list of rules which I can replace on
>"obj" later to find the maximum. No problem here*)
>rules = Table[
>FindRoot[{cpoAg1 == 0, cpoAg2 == 0}, {{x, 0.1}, {y, 0.1}}], {a, 0.=
1,
>1, 0.1}
>];
>(*the maximum is found with this following function *)
>Max[Thread[ReplaceAll[obj, rules]]];
>So far so good...I could discretise "a" as thinly as I want, but I
>can't afford the luxury of being that inneficient, 'cause I've got
>to do other things later with this part of the program.
I don't understand why you would take this approach. Why not use
the built-in function NMaximize, That is:
In[20]:= NMaximize[{obj,
cpoAg1 == 0 && cpoAg2 == 0 && x > 0.1 && y > 0.1}, {a, x, y}]
Out[20]= {2084.72,{a->0.527688,x->22.6727,y->13.7173}}
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