2220.11 – Timbuktu Post Office


Tags:Problem Set 9Algebra/TrigonometryAlgebra

When a careful count was taken a few years back, it was found that the Timbuktu Post Office handled at least 3,120 letters and 1,890 advertising brochures daily. The canceling machine then in use could handle 390 letters and 315 brochures per hour at the cost of $2.10 per hour. The post office had plans to acquire a new and improved model, which would be able to handle 520 letters and 210 brochures per hour at the cost of $2.70 per hour. Neither machine can be run for more than 10 hours each day or handle more than the maximum number of letters or brochures in an hour, or it will overheat and break down. Bearing these constraints in mind, what would be the most economically sound way to run the post office if they were to purchase the new machine? Note that the new machine will be in addition to the old machine, not a replacement for it.


Solution

The old machine can handle the workload by itself by running for 8 hours each day, at a cost of $16.80 per day.

Note that both of the two machines cancel letters at a rate of $0.005 per letter. However, the old machine cancels brochures at a rate of $0.007 per brochure, and the new machine cancels brochures at a rate of $0.013 per brochure. If we run the old machine long enough to cancel all the brochures—6 hours—we still have 780 letters to cancel:

3,1206(390)=780.3,120-6(390)=780.

Since the new machine works faster on letters, it will only take the new machine 1.5 hours to cancel the remaining letters. The total cost, therefore, is:

6($2.10)+1.5($2.70)=$16.65.6(\text{\textdollar}2.10)+1.5(\text{\textdollar}2.70)=\text{\textdollar}16.65.

This means that the Timbuktu Post Office will save $0.15 each day. Whether this is worth the investment is another question. It is best for the post office to use the old machine until all brochures are canceled, and then use the new machine for any remaining letters.