New Year's Countdown Day 19: Studying for Finals

Quantitative Finance Level 2

Justin has to study for final exams, but he'd rather be on Brilliant doing problems. He has 19 hours before he has to attend school to take his finals, so he will spend as much of that time on studying and completing quizzes on Brilliant. He compiles these two charts that show how much utility (satisfaction) he would gain from studying, and how much utility he would gain from completing quizzes:

Number of exams studied for 1 2 3 4 5 6 Total utility 19 34 48 60 70 78 \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Number of exams studied for} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{Total utility} & 19 & 34 & 48 & 60 & 70 & 78 \\ \hline \end{array}

Number of quizzes completed 1 2 3 4 5 6 7 8 9 Total utility 19 37 53 67 79 89 96 100 101 \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{Number of quizzes completed} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text{Total utility} & 19 & 37 & 53 & 67 & 79 & 89 & 96 & 100 & 101 \\ \hline \end{array}

Given that it takes Justin 3 hours to study for each exam and 2 hours to complete each quiz on Brilliant, how many exams should Justin study for in order to get as much utility as possible?

Details and Assumptions:

  • Assume that it is not possible to study for a fraction of an exam or complete a fraction of a quiz.
  • Assume that Justin will only study and do quizzes for as long as possible.
  • Justin gains no utility from either studying for zero exams or completing zero quizzes.
This problem is part of the set New Year's Countdown 2017 .
1 3 Not enough information to conclude 5 6 0 2 4

Steven Yuan by Steven Yuan , 20, USA

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1 solution

Steven Yuan
Dec 19, 2017

Let E = number of exams studies for and Q = number of quizzes completed on Brilliant. Since we are maximizing utility, we want to first find the marginal utility per hour for both activities, which we can do by first finding the marginal utility of each additional unit of each activity, then dividing by the time it takes to complete each activity (3 for exams, 2 for quizzes):

E TU E MU E MU E /hr 1 19 19 6.33 2 34 15 5 3 48 14 4.67 4 60 12 4 5 70 10 3.33 6 78 8 2.67 \begin{array}{|c|c|c|c|} \hline \text{E} & \text{TU}_{\text{E}} & \text{MU}_{\text{E}} & \text{MU}_{\text{E}} \text{/hr} \\ \hline 1 & 19 & 19 & 6.33 \\ 2 & 34 & 15 & 5 \\ 3 & 48 & 14 & 4.67 \\ 4 & 60 & 12 & 4 \\ 5 & 70 & 10 & 3.33 \\ 6 & 78 & 8 & 2.67 \\ \hline \end{array}

Q TU Q MU Q MU Q /hr 1 19 19 9.5 2 37 18 9 3 53 16 8 4 67 14 7 5 79 12 6 6 89 10 5 7 96 7 3.5 8 100 4 2 9 101 1 0.5 \begin{array}{|c|c|c|c|} \hline \text{Q} & \text{TU}_{\text{Q}} & \text{MU}_{\text{Q}} & \text{MU}_{\text{Q}} \text{/hr} \\ \hline 1 & 19 & 19 & 9.5 \\ 2 & 37 & 18 & 9 \\ 3 & 53 & 16 & 8 \\ 4 & 67 & 14 & 7 \\ 5 & 79 & 12 & 6 \\ 6 & 89 & 10 & 5 \\ 7 & 96 & 7 & 3.5 \\ 8 & 100 & 4 & 2 \\ 9 & 101 & 1 & 0.5 \\ \hline \end{array}

Now, we "check off" each unit of activity based on which side has the greater marginal utility per hour:

  1. We check off the first four units of quizzes because the MU/hour for each unit is greater than 6.33, which is the MU/hour for the first unit of exams.
  2. Because 6.33 is greater than 6 (the MU/hour of the fifth unit of quizzes), we check off the first unit of exams.
  3. Because 6 is greater than 5 (the MU/hour of the second unit of exams), we check off the fifth unit of quizzes.
  4. Because both sides have MU/hour equal to 5, we check off both sides: the second unit of exams and the sixth unit of quizzes.

At this point, we have used up 3 ( 2 ) + 2 ( 6 ) = 6 + 12 = 18 3(2) + 2(6) = 6 + 12 = 18 hours of time. Since we cannot have fractional quantities of either activity, this is as far as we can go, so Justin should study for only 2 \boxed{2} exams in order to maximize his total utility.

(And yes, I am in the midst of finals this week, and I also have college apps to do, so I haven't been able to post these countdown problems on time. But once finals are over, I will be back full-time on Brilliant!)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

But for 2 exams and 6 quizzes he gets 34 + 89 = 123 units of utility in your scheme.

If he does 3 exams and 5 quizzes he would get 48 + 79 = 127 units of utility. This seems more optimal, right?

Steven Perkins - 3 years, 5 months ago

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Hmm, you're right actually. I may have to rewrite this problem if it doesn't get reported first. But first I'm gonna check the solution and see where the work went wrong

Steven Yuan - 3 years, 5 months ago

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