<p>Another way to represent the preference is:</p>
<p><span class="math-container">\begin{eqnarray*} u(x_1, x_2, x_3) &amp; = &amp; x_1 + x_2 + x_3 - \min(x_1, x_2, x_3) \\ &amp; = &amp; \max(x_2+x_3, x_1+x_3, x_1+x_2)\end{eqnarray*}</span>
You can divide the above by <span class="math-container">$2$</span> to write the utility in terms of average.</p>
<p>Given that <span class="math-container">$x_1=60$</span>,
<span class="math-container">\begin{eqnarray*} u(60, x_2, x_3) &amp; = &amp; 60 + x_2 + x_3 - \min(60, x_2, x_3) \\ &amp; = &amp; \max(x_2+x_3, 60+x_3, 60+x_2)\end{eqnarray*}</span></p>
<p>Now <span class="math-container">$u(60, 50, 70) = 130$</span> (sum of the top two scores). So the indifference curve where <span class="math-container">$u = 130$</span> is the set:
<span class="math-container">$\{(x_1, x_2, x_3)\in\mathbb{R}^3_+| \max(x_2+x_3, x_1+x_3, x_1+x_2) = 130\}$</span></p>
<p>And indifference curve holding <span class="math-container">$x_1 =60$</span> fixed is:
<span class="math-container">$\{(x_2, x_3)\in\mathbb{R}^2_+| \max(x_2+x_3, 60+x_3, 60+x_2) = 130\}$</span></p>
<p>Observe that <span class="math-container">$\{(x_2, x_3)\in\mathbb{R}^2_+| \max(x_2+x_3, 60+x_3, 60+x_2) = 130\}$</span> <span class="math-container">$\subset \{(x_2, x_3)\in\mathbb{R}^2_+| x_2+x_3 = 130 \vee x_3 = 70 \vee x_2 = 70\}$</span></p>
<p>So we can first plot the lines <span class="math-container">$x_2+x_3 = 130$</span>, <span class="math-container">$x_3 = 70$</span>, <span class="math-container">$x_2 = 70$</span> and get:</p>
<p><a href="https://i.sstatic.net/Eznta.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Eznta.png" alt="enter image description here" /></a></p>
<p>Now we can eliminate the part of these lines where <span class="math-container">$\max(x_2+x_3, 60+x_3, 60+x_2) &gt; 130$</span> and get the indifference curve:</p>
<p><a href="https://i.sstatic.net/ITHkx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ITHkx.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-05-05T09:55:42.300Z
2022-05-05T09:55:42.300Z
Amit
<p>The Utility function here is the average score that Polly gets i.e.</p>

<p>$U(x_2,x_3)=\frac{max\{x_2,x_3,60\}+max^2\{x_2,x_3,60\}}{2}\\$</p>

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

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

<p>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 $(2c-60,60)$.</p>

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

CC BY-SA 3.0
2015-04-19T15:59:17.193Z
2015-04-19T15:59:17.193Z
Dreyfus
<p>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. </p>

<p>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 first 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 indifference 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 indifference
curve through the point ($x_2; x_3$) = (50; 70) is:</p>

<ol>
<li>L-shaped with a kink where $x_2 = x_3$.</li>
<li>three line segments, one vertical, one horizontal, and one running from (70; 60) to (60; 70).</li>
<li>a straight line, running from (0; 120) to (120; 0).</li>
<li>three line segments, one vertical, one horizontal, and one running from (70; 50) to (50; 70).</li>
<li>a V-shaped curve with its point at (50; 70).</li>
</ol>

<microeconomics><preferences>
Deriving indifference curves
dexter
# Deriving indifference curves

(Asked by: username_0 on 2015-04-19)

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 first 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 indifference 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 indifference
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).

**Tags:** microeconomics, preferences


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## 2 answers

### Answer by username_2 on 2015-04-19. Score: 3 | Accepted answer

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

$U(x\_2,x\_3)=\frac{max\{x\_2,x\_3,60\}+max^2\{x\_2,x\_3,60\}}{2}\\$

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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### Answer by username_1 on 2022-05-05. Score: 2

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](https://i.sstatic.net/Eznta.png)](https://i.sstatic.net/Eznta.png)

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](https://i.sstatic.net/ITHkx.png)](https://i.sstatic.net/ITHkx.png)


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