Sunday

August 2, 2015

August 2, 2015

Posted by **Alicia** on Saturday, October 4, 2008 at 11:53am.

- 11 grade math -
**David Q**, Saturday, October 4, 2008 at 2:44pmMS = Moe Singles; MD = Moe Doubles;

LS = Larry Singles; LD = Larry Doubles;

CS = Curlie Singles; CD = Curlie Doubles.

Then:

MS = 3 x LS

LD = 4 x CD

MS + (2 x MD) = LS + (2 x LD) = CS + (2 x CD)

MS + LS + CS = MD + LD + SD

MS + LS + CS + 2 x (MD + LD + SD) < 200.

It seems to me that there ought to be at least one more condition, because as it stands there appears to be a trivial solution of everyone having zero singles and zero doubles. I'm not familiar with the rules of baseball however, so it may be that there's something I need to know about the minimum number of hits possible for any one player. Or is the question actually that the total number of hits must be as large as possible, subject to the constraint that Total < 200?

If so, there's almost a solution with a total of 204 hits at:

MS=12, LS=4, CS=52, MD=28, LD=32, CD=8

All of these numbers are divisible by 4, so there's also a solution with a total of 3x204/4=153 hits at:

MS=9, LS=3, CS=39, MD=21, LD=24, CD=6, which I suspect is probably the highest possible, but I haven't been able to prove that. Obviously there will be two more solutions corresponding to one-half and one-quarter of the total of 204 hits.

How do you find them? Feed it though a linear programming package or MS Excel's "Solver" add-in, and you'll get the optimal solution (for Total=200) is:

MS=11.76, LS=3.92, CS=50.98, MD=27.45, LD=31.37, CD=7.84.

Then tweak the above until all of the numbers are integers. If there's an analytical way of finding it I've no idea what it might be, so if you're expected to do it that way, all I can supply is the answer against which to check your method.