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).

2 Answers 2


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.


Another way to represent the preference is:

\begin{eqnarray*} u(x_1, x_2, x_3) & = & x_1 + x_2 + x_3 - \min(x_1, x_2, x_3) \\ & = & \max(x_2+x_3, x_1+x_3, x_1+x_2)\end{eqnarray*} You can divide the above by $2$ to write the utility in terms of average.

Given that $x_1=60$, \begin{eqnarray*} u(60, x_2, x_3) & = & 60 + x_2 + x_3 - \min(60, x_2, x_3) \\ & = & \max(x_2+x_3, 60+x_3, 60+x_2)\end{eqnarray*}

Now $u(60, 50, 70) = 130$ (sum of the top two scores). So the indifference curve where $u = 130$ is the set: $\{(x_1, x_2, x_3)\in\mathbb{R}^3_+| \max(x_2+x_3, x_1+x_3, x_1+x_2) = 130\}$

And indifference curve holding $x_1 =60$ fixed is: $\{(x_2, x_3)\in\mathbb{R}^2_+| \max(x_2+x_3, 60+x_3, 60+x_2) = 130\}$

Observe that $\{(x_2, x_3)\in\mathbb{R}^2_+| \max(x_2+x_3, 60+x_3, 60+x_2) = 130\}$ $\subset \{(x_2, x_3)\in\mathbb{R}^2_+| x_2+x_3 = 130 \vee x_3 = 70 \vee x_2 = 70\}$

So we can first plot the lines $x_2+x_3 = 130$, $x_3 = 70$, $x_2 = 70$ and get:

enter image description here

Now we can eliminate the part of these lines where $\max(x_2+x_3, 60+x_3, 60+x_2) > 130$ and get the indifference curve:

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.