I believe there is some good benchmarking testing http://12000.org/my_notes/rankTest/test.htm

Nice work!

You don't appear to have Maple 9 in your repetoire.  Maple 9.5 was the next major release one year later from Maple 9.  I don't think there's much of a difference between the two versions though.

@mithun helllooooo  .... did you you try clearing the autoexecute or changing the startup sheet like I suggested? 

Are you suggesting the startup screen?

In this solution ..

BinPacking1D(8*12, [[34.5, 2], [96, 2], [8, 2], [22, 2], [40, 2]]);

 [[96] = 2, [22, 22] = 1, [34.5, 34.5] = 1, [8, 8, 40, 40] = 1]

Grouping one of the 22's with 34.5 would give us a longer left over piece, on the same token we could move a 34.5 over to the 22" group.  But what is optimal?

Is it better to have 52" and 69"  or  5" and 74"  or  17.5" and 61.5"  left over?  Certainly we would get the most use out of a 74" piece but the 5" is pretty much useless. However the original solution giving us 52" and 69" would potentially leave us with two useful pieces as opposed to only one.

Strangely a 1/4" smaller piece gives us quite a different ordering.

BinPacking1D(8*12, [[34.25, 2], [96, 2], [8, 2], [22, 2], [40, 2]]);

[[96] = 2, [40, 40] = 1, [8, 34.25] = 1, [8, 22, 22, 34.25] = 1]

leaving us with 16", 53.75" and 9.75"

Interestingly a length of 34.75" offers the same solution as that with 34.25".  Hmm, maybe the decimal has something to do with the outcome of  the solution?
So just for fun I thought to try 34.50 instead of 34.5 and wouldnt' you know it the returned solution was the same as for 34.75 and 34.25. 

What is it about using 34.5 instead of 34.50 that changes the solution?

ah!  I forgot to add a:=[] before the dsolve, but yes even then I wasn't allowed := inside the [trigger,action] pair.  I thought also of using assign in the [trigger,action] pair but it failed to work as I wanted there as well. 

I think I see how it is working in your above example, it's a little more involved than I originally thought.

@Preben Alsholm

is it possible to add an events triggerand compile it into a list in dsolve.

problem is I get and invalid sequence error
sol := dsolve({EL, ini},numeric,events=[[theta(t)=0,diff(theta(t),t)=-diff(theta(t),t)],[diff(theta(t),t)=0,a:=[op(a),t]]]);

How do I find at what value of t the maximums of theta occur?

@Preben Alsholm ok that's much better :) thanks

If I set U=-mgRcos(theta) then the Lagrangian ends up being

which appears to work perfectly instead of setting U=+mgRcos(theta) which the Lagrangian ends up as

as shown in textbooks but graphically doesn't appear correct?

Maybe it's the naming convention I'm getting confused on but this is just confusing me to the nth degree at the moment. 

Please shed a little more light on the subject for me.

@Preben Alsholm  no offense intended.  However your question made me think.  In Marion and Thornton the sign on U is positive in a similar problem so I just followed the same convention.  So should it be negative here in my question?

@Preben Alsholm thanks!  Maybe the sign is wrong, the potential energy at the initial point is positive so that is why I have made U positive.  In a free body diagram yes I would label the force as negative but I don't think it is negative here.  

Thanks about the events I did not know about events.  Again not near Maple till this evening.

So if it were a bead on a frictionless wire and hit the ground with a completely inelastic collision I would just do

sol := dsolve({EL, ini},numeric,events=[[theta(t)=0,diff(theta(t),t)=0]]);

Also if it were an elastic collision and I wanted to include some dissipative terms (ie friction or coefficient of restitution) I would have to add Raleigh dissipative terms in the Lagrangian L=T-U+R I know off hand the friction term should fit nicely (ie with respect to velocity of the bead) but haven't thought much about the coefficient of restitution part and I'm not sure actually how I would add that part in there at the moment.

Sorry for the delay, didn't have access to Maple until now.  Strangely I could not replicate the D(theta)(0)=0 becoming 0=0 issue.  It must have been something I did unknowingly.

Curiously I was working on a constrained particle.  I was working on a graphical solution.  I suppose one must know when the particle hits the ground as the graph acts as it just continuously moves in a circle. 

Any ideas for the ground contraints?  I have attached what I have worked on here.

Download constrained_lagrangian.mw

@fairymath The application is here, in the application center.

Yes I agree, his applications are always worth looking forward to.

How many versions back does this issue go?