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I have been trying to work this out for quite a while now, but I can't seem to understand how to solve these kind of questions. Any help (or hint) would be highly appreciated.

Professor Goodheart's colleague Dr. Kremepu gives 3 midterm exams. He drops the lowest and gives each student her average score on the other two exams. Polly Sigh is taking his course and has a 60 on her fi rst exam. Let $x_2$ be her score on the second exam and $x_3$ be her score on the third exam. If we draw her indiff erence curves for scores on the second and third exams with $x_2$ represented by the horizontal axis and $x_3$ represented by the vertical axis, then her indi fference curve through the point ($x_2; x_3$) = (50; 70) is:

  1. L-shaped with a kink where $x_2 = x_3$.
  2. three line segments, one vertical, one horizontal, and one running from (70; 60) to (60; 70).
  3. a straight line, running from (0; 120) to (120; 0).
  4. three line segments, one vertical, one horizontal, and one running from (70; 50) to (50; 70).
  5. a V-shaped curve with its point at (50; 70).
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up vote 2 down vote accepted

The Utility function here is the average score that Polly gets i.e.


Where: $max^2\{.\}$ stands for second highest number. Now this utility function can be rewritten as:

$U(x_2,x_3)= \begin{cases}\frac{x_2+x_3}{2}, & \text{ if } x_2,x_3\geq 60,\\ \frac{60+x_3}{2}, & \text{ if } x_2<60, x_3>x_2,\\ \frac{60+x_2}{2}, & \text{ if } x_3<60,x_3<x_2,\\ \frac{60+k}{2}, & \text{ if } k<60,x_3=x_2=k.\\ \end{cases}$

Plotting its level curve i.e. the IC for utility level $c$ would be a horizontal line from point $(0,2c-60)$ to $(60,2c-60)$, a vertical line from point $(2c-60,0)$ to $(60,2c-60)$ and a line from $(60,2c-60)$ to $(2c-60,60)$.

So if $(x_2,x_3)=(50,70)$ option 2 satisfies the IC's description.

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