<p>We need to solve
<span class="math-container">$$
\max_{x_1, x_2} \left(\max\{\min\{x_1, 2 x_2\}, \min\{x_2, 2 x_1\}\}\right) \text{ s.t. } p_1 x_1 + p_2 x_2 \le m
$$</span></p>
<ol>
<li><p>assume <span class="math-container">$x_1 \le 2 x_2$</span> and <span class="math-container">$x_2 \le 2 x_1$</span>, then:
<span class="math-container">$$
u(x_1, x_2) = \max\{x_1, x_2\}.
$$</span>
1.1 Now if <span class="math-container">$x_1 \ge x_2$</span> then this gives <span class="math-container">$u(x_1, x_2) = x_1$</span>. We want to maximize this, so we put <span class="math-container">$x_1$</span> as high as possible, this gives <span class="math-container">$x_1 = 2 x_2$</span>. Substituting into the budget constraint gives <span class="math-container">$x_2 = \frac{m}{2 p_1 + p_2}$</span> and <span class="math-container">$x_1 = \frac{2m}{2p_1 + p_2}$</span>. Notice that indeed <span class="math-container">$x_2 \le 2 x_1$</span>. The utility obtained is <span class="math-container">$u(x_1, x_2) = \frac{2m}{2p_1 + p_2}$</span>.</p>
<p>1.2 If <span class="math-container">$x_2 \ge x_1$</span> then this gives <span class="math-container">$u(x_1, x_2) = x_2$</span>. We want to maximize this, so we put <span class="math-container">$x_2$</span> as high as possible. This gives <span class="math-container">$x_2 = 2 x_1$</span>. Substituting into the budget constraint gives <span class="math-container">$x_1 = \frac{m}{p_1 + 2 p_2}$</span> and <span class="math-container">$x_2 = \frac{2m}{p_1 + 2 p_2}$</span>. Then <span class="math-container">$u(x_1, x_2) = \frac{2m}{p_1 + 2 p_2}$</span>.</p>
</li>
<li><p>if <span class="math-container">$x_1 \le 2 x_2$</span> and <span class="math-container">$2 x_1 \le x_2$</span> (notice that only the latter restriction is really necessary). Then:
<span class="math-container">$$
u(x_1, x_2) = \max\{x_1, 2x_1\} = 2 x_1
$$</span>
Here, we want to make <span class="math-container">$x_1$</span> as large as possible. This gives <span class="math-container">$2 x_1 = x_2$</span>. Substituting into the budget constraint gives <span class="math-container">$x_1 = \frac{m}{p_1 + 2 p_2}$</span> and <span class="math-container">$x_2 = \frac{2m}{p_1 + 2 p_2}$</span>. The utility is given by <span class="math-container">$u(x_1, x_2) = \frac{2m}{p_1 + 2p_2}$</span>.</p>
</li>
<li><p>if <span class="math-container">$2 x_2 \le x_1$</span> and <span class="math-container">$x_2 \le 2 x_1$</span>. (notice that only hte first restriction is needed). Then
<span class="math-container">$$
u(x_1, x_2) = \max\{2 x_2, x_2\} = 2 x_2.
$$</span>
Here, to make <span class="math-container">$x_2$</span> as large as possible, we will have <span class="math-container">$2 x_2 = x_1$</span>. Then substituting into the budget constraint gives <span class="math-container">$x_2 = \frac{m}{2p_1 + p_2}$</span>. and <span class="math-container">$x_1 = \frac{2m}{2p_1 + p_2}$</span>. The utility is given by <span class="math-container">$u(x_1, x_2) = \frac{2m}{2p_1 + p_2}$</span>.</p>
</li>
<li><p>If <span class="math-container">$2x_2 \le x_1$</span> and <span class="math-container">$2 x_1 \le x_2$</span>. Then <span class="math-container">$2 x_2 \le x_1 \le 2 x_1 \le x_2$</span>, which is possible only if <span class="math-container">$x_2 = 0$</span> and <span class="math-container">$x_1 = 0$</span>, but this is never utility maximising.</p>
</li>
</ol>
<p>Summarizing, we can choose between <span class="math-container">$u(x_1, x_2) = \frac{2m}{2p_1 + p_2}$</span> or <span class="math-container">$u(x_1, x_2) = \frac{2m}{p_1 + 2 p_2}$</span>. The first will be the optimal choice (give more utility) if <span class="math-container">$p_1 &lt; p_2$</span> the second will be optimal if <span class="math-container">$p_1 &gt; p_2$</span>. If <span class="math-container">$p_1 = p_2$</span> we are indifferent.</p>

CC BY-SA 4.0
2023-09-01T09:30:48.023Z
2023-09-01T09:30:48.023Z
tdm
<p>More generally, if you want to solve the problem:
<span class="math-container">\begin{eqnarray*} \max_{(x_1,x_2,\dots, x_n)\in C} \max(f_1(x_1,x_2,\ldots, x_n), f_2(x_1,x_2,\ldots, x_n),\ldots, f_m(x_1,x_2,\ldots, x_n))\end{eqnarray*}</span>
Solve each of these <span class="math-container">$m$</span> problems:
<span class="math-container">\begin{eqnarray*} \max_{(x_1,x_2,\dots, x_n)\in C} f_j(x_1,x_2,\ldots, x_n) \end{eqnarray*}</span>
where <span class="math-container">$j\in\{1,2,\ldots,m\}$</span></p>
<p>Assuming the conditions for the existence of these <span class="math-container">$m$</span> solutions hold (we can assume <span class="math-container">$f_j$</span> is continuous and <span class="math-container">$C$</span> is a compact subset of <span class="math-container">$\mathbb{R}^n$</span>), let <span class="math-container">$(x_1^j,x_2^j,\dots, x_n^j)$</span> be one of the solutions to the problem
<span class="math-container">\begin{eqnarray*} \max_{(x_1,x_2,\dots, x_n)\in C} f_j(x_1,x_2,\ldots, x_n) \end{eqnarray*}</span>
so that <span class="math-container">$\max_{(x_1,x_2,\dots, x_n)\in C} f_j(x_1,x_2,\ldots, x_n)=f_j(x_1^j,x_2^j,\ldots, x_n^j)$</span></p>
<p>Proposition.<br />
<span class="math-container">\begin{eqnarray*} \max_{(x_1,x_2,\dots, x_n)\in C} \max(f_1(x_1,x_2,\ldots, x_n), f_2(x_1,x_2,\ldots, x_n),\ldots, f_m(x_1,x_2,\ldots, x_n))= \max_{j\in\{1,2,\ldots,m\}} f_j(x_1^j,x_2^j,\ldots, x_n^j)\end{eqnarray*}</span></p>
<p>and</p>
<p><span class="math-container">$j^*$</span> solves the problem
<span class="math-container">\begin{eqnarray*} \max_{j\in\{1,2,\ldots,m\}} f_j(x_1^j,x_2^j,\ldots, x_n^j)\end{eqnarray*}</span>
if and only if <span class="math-container">$(x_1^{j^*},x_2^{j^*},\ldots, x_n^{j^*})$</span> solves the problem <span class="math-container">\begin{eqnarray*} \max_{(x_1,x_2,\dots, x_n)\in C} \max(f_1(x_1,x_2,\ldots, x_n), f_2(x_1,x_2,\ldots, x_n),\ldots, f_m(x_1,x_2,\ldots, x_n))\end{eqnarray*}</span>
Proof. Easy</p>
<p>We can now use this proposition and solve the given problem easily.
To summarise, we simply change the order of optimization:
<span class="math-container">\begin{eqnarray*} \max_{(x_1,x_2,\dots, x_n)\in C} \max_{j\in\{1,2,\ldots,m\}} f_j(x_1,x_2,\ldots, x_n) = \max_{j\in\{1,2,\ldots,m\}} \max_{(x_1,x_2,\dots, x_n)\in C}  f_j(x_1,x_2,\ldots, x_n)\end{eqnarray*}</span></p>

CC BY-SA 4.0
2023-09-01T17:47:27.777Z
Very clever @Amit +1!
Nicolas Torres
CC BY-SA 4.0
2023-09-01T09:35:13.607Z
2023-09-01T09:42:39.497Z
2023-09-01T09:42:39.497Z
Amit
<p>I have a rather bizarre utility function, which is neither differentiable nor quasi-concave
<span class="math-container">$$u(x_1,x_2)=\max\{\min\{x_1,2x_2\},\min\{2x_1,x_2\}\}$$</span>
How to derive the demand function?</p>

CC BY-SA 4.0
2023-09-01T04:56:28.820Z
Refer to https://qr.ae/pvDf9H
Amit
<demand>
How to derive the demands from this rather bizarre utility?
Ludwig Gershwin
# How to derive the demands from this rather bizarre utility?

(Asked by: username_0 on 2023-09-01)

I have a rather bizarre utility function, which is neither differentiable nor quasi-concave
$$u(x\_1,x\_2)=\max\{\min\{x\_1,2x\_2\},\min\{2x\_1,x\_2\}\}$$
How to derive the demand function?

**Tags:** demand

> **Comments:**
> **username_1**: Refer to https://qr.ae/pvDf9H  

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## 2 answers

### Answer by username_2 on 2023-09-01. Score: 0 | Accepted answer

We need to solve
$$
\max\_{x\_1, x\_2} \left(\max\{\min\{x\_1, 2 x\_2\}, \min\{x\_2, 2 x\_1\}\}\right) \text{ s.t. } p\_1 x\_1 + p\_2 x\_2 \le m
$$

1. assume $x\_1 \le 2 x\_2$ and $x\_2 \le 2 x\_1$, then:
   $$
   u(x\_1, x\_2) = \max\{x\_1, x\_2\}.
   $$
   1.1 Now if $x\_1 \ge x\_2$ then this gives $u(x\_1, x\_2) = x\_1$. We want to maximize this, so we put $x\_1$ as high as possible, this gives $x\_1 = 2 x\_2$. Substituting into the budget constraint gives $x\_2 = \frac{m}{2 p\_1 + p\_2}$ and $x\_1 = \frac{2m}{2p\_1 + p\_2}$. Notice that indeed $x\_2 \le 2 x\_1$. The utility obtained is $u(x\_1, x\_2) = \frac{2m}{2p\_1 + p\_2}$.

   1.2 If $x\_2 \ge x\_1$ then this gives $u(x\_1, x\_2) = x\_2$. We want to maximize this, so we put $x\_2$ as high as possible. This gives $x\_2 = 2 x\_1$. Substituting into the budget constraint gives $x\_1 = \frac{m}{p\_1 + 2 p\_2}$ and $x\_2 = \frac{2m}{p\_1 + 2 p\_2}$. Then $u(x\_1, x\_2) = \frac{2m}{p\_1 + 2 p\_2}$.
2. if $x\_1 \le 2 x\_2$ and $2 x\_1 \le x\_2$ (notice that only the latter restriction is really necessary). Then:
   $$
   u(x\_1, x\_2) = \max\{x\_1, 2x\_1\} = 2 x\_1
   $$
   Here, we want to make $x\_1$ as large as possible. This gives $2 x\_1 = x\_2$. Substituting into the budget constraint gives $x\_1 = \frac{m}{p\_1 + 2 p\_2}$ and $x\_2 = \frac{2m}{p\_1 + 2 p\_2}$. The utility is given by $u(x\_1, x\_2) = \frac{2m}{p\_1 + 2p\_2}$.
3. if $2 x\_2 \le x\_1$ and $x\_2 \le 2 x\_1$. (notice that only hte first restriction is needed). Then
   $$
   u(x\_1, x\_2) = \max\{2 x\_2, x\_2\} = 2 x\_2.
   $$
   Here, to make $x\_2$ as large as possible, we will have $2 x\_2 = x\_1$. Then substituting into the budget constraint gives $x\_2 = \frac{m}{2p\_1 + p\_2}$. and $x\_1 = \frac{2m}{2p\_1 + p\_2}$. The utility is given by $u(x\_1, x\_2) = \frac{2m}{2p\_1 + p\_2}$.
4. If $2x\_2 \le x\_1$ and $2 x\_1 \le x\_2$. Then $2 x\_2 \le x\_1 \le 2 x\_1 \le x\_2$, which is possible only if $x\_2 = 0$ and $x\_1 = 0$, but this is never utility maximising.

Summarizing, we can choose between $u(x\_1, x\_2) = \frac{2m}{2p\_1 + p\_2}$ or $u(x\_1, x\_2) = \frac{2m}{p\_1 + 2 p\_2}$. The first will be the optimal choice (give more utility) if $p\_1 < p\_2$ the second will be optimal if $p\_1 > p\_2$. If $p\_1 = p\_2$ we are indifferent.


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### Answer by username_1 on 2023-09-01. Score: 3

More generally, if you want to solve the problem:
\begin{eqnarray\*} \max\_{(x\_1,x\_2,\dots, x\_n)\in C} \max(f\_1(x\_1,x\_2,\ldots, x\_n), f\_2(x\_1,x\_2,\ldots, x\_n),\ldots, f\_m(x\_1,x\_2,\ldots, x\_n))\end{eqnarray\*}
Solve each of these $m$ problems:
\begin{eqnarray\*} \max\_{(x\_1,x\_2,\dots, x\_n)\in C} f\_j(x\_1,x\_2,\ldots, x\_n) \end{eqnarray\*}
where $j\in\{1,2,\ldots,m\}$

Assuming the conditions for the existence of these $m$ solutions hold (we can assume $f\_j$ is continuous and $C$ is a compact subset of $\mathbb{R}^n$), let $(x\_1^j,x\_2^j,\dots, x\_n^j)$ be one of the solutions to the problem
\begin{eqnarray\*} \max\_{(x\_1,x\_2,\dots, x\_n)\in C} f\_j(x\_1,x\_2,\ldots, x\_n) \end{eqnarray\*}
so that $\max\_{(x\_1,x\_2,\dots, x\_n)\in C} f\_j(x\_1,x\_2,\ldots, x\_n)=f\_j(x\_1^j,x\_2^j,\ldots, x\_n^j)$

Proposition.  
\begin{eqnarray\*} \max\_{(x\_1,x\_2,\dots, x\_n)\in C} \max(f\_1(x\_1,x\_2,\ldots, x\_n), f\_2(x\_1,x\_2,\ldots, x\_n),\ldots, f\_m(x\_1,x\_2,\ldots, x\_n))= \max\_{j\in\{1,2,\ldots,m\}} f\_j(x\_1^j,x\_2^j,\ldots, x\_n^j)\end{eqnarray\*}

and

$j^\*$ solves the problem
\begin{eqnarray\*} \max\_{j\in\{1,2,\ldots,m\}} f\_j(x\_1^j,x\_2^j,\ldots, x\_n^j)\end{eqnarray\*}
if and only if $(x\_1^{j^\*},x\_2^{j^\*},\ldots, x\_n^{j^\*})$ solves the problem \begin{eqnarray\*} \max\_{(x\_1,x\_2,\dots, x\_n)\in C} \max(f\_1(x\_1,x\_2,\ldots, x\_n), f\_2(x\_1,x\_2,\ldots, x\_n),\ldots, f\_m(x\_1,x\_2,\ldots, x\_n))\end{eqnarray\*}
Proof. Easy

We can now use this proposition and solve the given problem easily.
To summarise, we simply change the order of optimization:
\begin{eqnarray\*} \max\_{(x\_1,x\_2,\dots, x\_n)\in C} \max\_{j\in\{1,2,\ldots,m\}} f\_j(x\_1,x\_2,\ldots, x\_n) = \max\_{j\in\{1,2,\ldots,m\}} \max\_{(x\_1,x\_2,\dots, x\_n)\in C} f\_j(x\_1,x\_2,\ldots, x\_n)\end{eqnarray\*}

> **Comments:**
> **username_3**: Very clever @username_1 +1!  

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