4. A decision at the margin
Carlos is a hard-working college senior. One Saturday, he decides to work nonstop until he has answered 100 practice problems for his physics course. He starts work at 8:00 AM and uses a table to keep track of his progress throughout the day. He notices that as he gets tired, it takes him longer to solve each problem.
Time | Total Problems Answered |
---|---|
8:00 AM | 0 |
9:00 AM | 40 |
10:00 AM | 70 |
11:00 AM | 90 |
Noon | 100 |
Use the table to answer the following questions.
The marginal, or additional, gain from Carlos’s second hour of work, from 9:00 AM to 10:00 AM, isproblems.
The marginal gain from Carlos’s fourth hour of work, from 11:00 AM to noon, isproblems.
Later, the teaching assistant in Carlos’s physics course gives him some advice. “Based on past experience,” the teaching assistant says, “working on 25 problems raises a student’s exam score by about the same amount as reading the textbook for 1 hour.” For simplicity, assume students always cover the same number of pages during each hour they spend reading.
Given this information, in order to use his 4 hours of study time to get the best exam score possible, how many hours should he have spent working on problems, and how many should he have spent reading?
1 hour working on problems, 3 hours reading
2 hours working on problems, 2 hours reading
3 hours working on problems, 1 hour reading
4 hours working on problems, 0 hours reading
Carlos solved 40 questions in 1st hour (8 to 9) then he solved 30 questions in 2nd hour (9 to 10) then he solved 20 questions in 3rd hour (10 to 11) and 10 questions in last 1 hour (11 to 12)
Carlos showed Diminishing returns over marginal and decreases by 10 questions each hour
Marginal Gain in 2nd Hour is 30 questions
Marginal Gain in 4th hour is 10 questions.
If Carlos solves problems for 1 hour i,e, 8 to 9 he will solve 40 problems and 3 hours reading is worth solving 3*25 =75 problems therefore total problems solved=115
If he solves for 2 hours he will solve 70 problems and 2 hours of reading will give 50 problems equivalent results therefore total problems solved =120
If he solves for 3 hours then total problems solved =90+25=105
Hence 2 hours working and 2 hours reading is the best choice
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