Calculators are allowed, but solving this without a computer is quite a challenge.
Here's a hint (rot13): Ubj zhpu jbhyq Cnfdhnyr or noyr gb znxr vs ur qvqa'g unir gur bcgvba bs hfvat cnegvnyyl-nffrzoyrq enj zngrevnyf?
This problem can be solved quite easily with a linear programming solver, such as the one at http://www.zweigmedia.com/RealWorld/simplex.html. Let p and q be the number of Ps and Qs produced each week using only raw materials, and let p' and q' be the number of Ps and Qs produced each week using partially-assembled raw materials.Click here to view the solution
Maximize z = 150p + 125p'+ 145q + 116q' subject to 18.5p + 16.5q + 15.5p' + 13q' <= 2400 p + p' <= 100 q + q' <= 100 Optimal Solution: z = 20581.1; p = 40.5405, p' = 0, q = 100, q' = 0So the most Pasquale can make is $20581.10/week. He can do this by avoiding the partially-assembled raw materials entirely, making as many Qs as the market will accept (100), and spending the remainder of his time on Ps (40.5405).
This puzzle was inspired by a paper by Balakrishnan .
 Balakrishnan, Jaydeep. The theory of constraints and the make-or-buy decision. Journal of Supply Chain Management, January 1, 2005 .
Post a Comment