Hi, I know this has been asked ( and answered before) but when I installed this package and pressed ?globalsearch the help page is void.how do I get it?

bags.mw

$15 is spread among four different bags. How  can you can make any whole number of dollars from $1 to $15 by selecting some of these bags of money, without taking any money out of the bags.
 
How is the money distributed ?

One bag must have $1 in it.
To be able to make $2, and each bag is different, so one bag must have $2 in it.
 can now make $3, but can’t make $4, so the third bag must have $4 in it.
With $1, $2 and $4 in three bags, can make every total up to $7. The last bag has the rest of his money in it – $8.

so the answer is $[1,2,4,8]

wondering how to do this with maple procedure. unsurprisingly my code doesn't work

This problem has real world applicability: Three vampires and three maidens are at the foot of a tall building and wish to get to the bar on the top floor.  The lift only holds two people (for convenience I am classing vampires as people), and needs one person to operate it.  If ever the vampires outnumber the maidens at any place, they will do something unspeakable.  How can the vampires and maidens all safety get to the top in the minimum of moves?  http://tonysmaths.blogspot.com.au/2012/05/vampires-and-maidens-problem.html

can this be solved procedurally or using optimization package?

possible manual soln:

No. m>= No. v

3m+3v, 0   [0]

2m+2v,m+v [1]

3m+2v,v  [2]

     3m, 3v [3]

3m+v , 2v  [4]

m+v  , 2m+2v [5]

#then reverse the steps

2m+2v, m+v [6]

       2v, 3m+v [7]

       3v, 3m  [8]

        v , 3m+2v [9]

    m+v, 2m+2v [10]

        0, 3m+3v [11]

Hello.

given this expression

T:=unapply((1/6930)*exp(-(1/7938)*(X[4]-933)^2)*exp(-(1/6050)*(X[2]-805)^2)/((1+exp((1/50)*X[4]-(1/50)*X[2]))*Pi),X[2]);

U := unapply(sum(T(X[2]), X[4] = 0 .. 3600), X[2]):

I want to display U, but not all 3600 terms. is there anyway to simplify/reduce this sum?

kind of like geo series a+ar+ar^2+ar^3+...+ar^(n-1)=sum(ar^k,k=0..n-1) can be reduced to a*(1-r^n)/(1-r)

given the following:

param:= {mu[2] = (1600), mu[4] = (1500)}
R := unapply(1/(1+exp((1/50)*x)), x);

prior := evalf(expand(eval(eval(R(abs(mu[4]-mu[2])), `` = (x->x)), param)));
   

I need to keep the paramaters encapsulated. I would like this last step to output the answer 0.1192029.... but the abs function doesn't let me