<p>We solve the problem</p>
<p><span class="math-container">\begin{equation}
    \max U(x_1,x_2,x_3) = \min\{x_1,x_2\} + x_3
\end{equation}</span></p>
<p>subject to</p>
<p><span class="math-container">\begin{equation}
    x_1 p_1 + x_2 p_2 + x_3 p_3 = I
\end{equation}</span></p>
<p>From the min term we get that <span class="math-container">$x_1 = x_2$</span>. Therefore the budget constraint becomes</p>
<p><span class="math-container">\begin{equation}
    x_1 (p_1 + p_2) + x_3 p_3 = I
\end{equation}</span></p>
<p>Solving for <span class="math-container">$x_1$</span> we get</p>
<p><span class="math-container">\begin{equation}
    x_1 = \frac{I - x_3 p_3}{p_1 + p_2}
\end{equation}</span></p>
<p>Since <span class="math-container">$x_1 = x_2$</span>,</p>
<p><span class="math-container">\begin{equation}
    x_2 = \frac{I - x_3 p_3}{p_1 + p_2}
\end{equation}</span></p>
<p>Therefore, the min term becomes</p>
<p><span class="math-container">\begin{equation}
    \min\{x_1,x_2\} = \frac{I - x_3 p_3}{p_1 + p_2}
\end{equation}</span></p>
<p>Substituting this expression for the min term into the utility function we get</p>
<p><span class="math-container">\begin{equation}
    W(x_3) := U(x_1(x_3),x_2(x_3),x_3) = \frac{I - x_3 p_3}{p_1 + p_2} + x_3
\end{equation}</span></p>
<p>Separating the fraction and rearranging we get</p>
<p><span class="math-container">\begin{equation}
    W(x_3) = (1 - \frac{p_3}{p_1 + p_2}) x_3 + \frac{I}{p_1 + p_2}
\end{equation}</span></p>
<p>Note that this is a straight line, with slope <span class="math-container">$1 - \frac{p_3}{p_1 + p_2}$</span>.</p>
<p>Notice <span class="math-container">$1 - \frac{p_3}{p_1 + p_2} &gt; 0 \iff 1 &gt; \frac{p_3}{p_1 + p_2} \iff p_1 + p_2 &gt; p_3$</span>.</p>
<p>From here we get 3 cases:</p>
<p><a href="https://i.sstatic.net/PEzjH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/PEzjH.png" alt="enter image description here" /></a></p>
<ul>
<li><span class="math-container">$p_1 + p_2 &gt; p_3 \rightarrow$</span> spend everything on <span class="math-container">$x_3 \rightarrow x_1 = 0, x_2 = 0, x_3 = \frac{I}{p_3}$</span>.</li>
<li><span class="math-container">$p_1 + p_2 &lt; p_3 \rightarrow$</span> don't consume <span class="math-container">$x_3 \rightarrow x_1 = \frac{I}{p_1 + p_2}, x_2 = \frac{I}{p_1 + p_2}, x_3 = 0$</span></li>
<li><span class="math-container">$p_1 + p_2 = p_3 \rightarrow x_3$</span> value indifferent <span class="math-container">$\rightarrow x_1 = \frac{I - x_3 p_3}{p_1 + p_2}, x_2 = \frac{I - x_3 p_3}{p_1 + p_2}, 0 \leq x_3 \leq \frac{I}{p_3}$</span>, i.e. the optimal bundles form a line in 3-D space that looks like this:</li>
</ul>
<p><a href="https://i.sstatic.net/vuCBD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vuCBD.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-11-14T01:49:41.130Z
2022-11-14T01:49:41.130Z
Nicolas Torres
<p>for this problem you must consider two possible branches of the utility function:</p>
<p><span class="math-container">$$u(\text{x})=x_1+x_3\ \ \text{if} \ \ x_1&lt;x_2$$</span>
<span class="math-container">$$u(\text{x})=x_2+x_3\ \ \text{if} \ \ x_1&gt;x_2$$</span></p>
<p>The demand equations you solve from each one of these provide guidance on what the solution would be to your problem. However you must list them for each case.</p>
<p>Hope this helps</p>

CC BY-SA 4.0
2019-11-10T04:22:24.773Z
2024-02-12T01:31:01.823Z
2024-02-12T01:31:01.823Z
EconJohn
<p>Hint: Imagine that there are two coffee bars, <span class="math-container">$A$</span> and <span class="math-container">$B$</span>. There is only one type of coffee in the world. My preferences are such that I always want 1 unit of sugar with 1 unit of coffee; if I consume units of coffee and sugar in the ratio <span class="math-container">$1:1$</span>, additional units of only one of the two don't give me extra utility.</p>

<p>At coffee bar <span class="math-container">$A$</span> they sell coffee and sugar separately. That is: there is one counter for coffee, where coffee is sold for a price of <span class="math-container">$p_1$</span> per unit. There is another counter for sugar, where I can buy one unit of sugar for a price of <span class="math-container">$p_2$</span>.</p>

<p>At coffee bar <span class="math-container">$B$</span> however, they only sell one unit of coffee with one unit of sugar in a bundle, the price of which is equal to <span class="math-container">$p_3$</span>. They do not sell anything separately</p>

<p>At what prices do I prefer to buy my coffee with sugar at <span class="math-container">$A$</span>, at what prices do I prefer to buy my coffee with sugar at <span class="math-container">$B$</span>?</p>

CC BY-SA 4.0
2019-10-09T18:55:06.407Z
If I understand you correctly, you buy at bar $A$ for $p1+p2$, and at bar $B$ for $p3$.
Anne1005
CC BY-SA 4.0
2019-10-09T19:36:46.790Z
That is correct, but what I meant with the last sentence was: At what values of $p_1$, $p_2$, $p_3$ is it cheaper to buy my coffee with sugar at $A$?
user18214
CC BY-SA 4.0
2019-10-09T19:39:33.527Z
I edited the answer to make it more clear, try to relate the little story above to the  problem you have to solve
user18214
CC BY-SA 4.0
2019-10-09T20:09:36.183Z
Thank you. In that case it means that you will buy at $A$ if $p1 + p2$ < $p3$. If $p1 + p2$ > $p3$, you will buy at $B$. If they are equal, you are indifferent.
Anne1005
CC BY-SA 4.0
2019-10-09T21:09:00.917Z
Exactly! And if you now combine this with the fact that your budget is equal to $w$ you have your demand function.
user18214
CC BY-SA 4.0
2019-10-09T15:39:34.263Z
2019-10-09T19:38:12.527Z
2019-10-09T19:38:12.527Z
user18214
<p>Since you already know how to solve for the two goods case with <span class="math-container">$u(x)=\min(x_1,x_2)$</span> you can solve the optimization problem in a two-stage method without using Lagrangian.</p>
<p>UMP: <span class="math-container">$$\begin{aligned}
\max_{x_1,x_2,x_3\geq 0} \quad &amp; \min(x_1,x_2)+x_3\\
\textrm{s.t.} \quad &amp; p_1x_1+p_2x_2+p_3x_3\leq M\end{aligned}$$</span></p>
<p>Stage one: Solving the problem with respect to <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> holding <span class="math-container">$x_3$</span> constant
<span class="math-container">$$\max_{x_3}\begin{pmatrix}{\begin{aligned}\max_{x_1,x_2} \quad &amp; \min(x_1,x_2)+x_3\\
\textrm{s.t. } \quad &amp; p_1x_1+p_2x_2\leq M-p_3x_3\end{aligned}}\end{pmatrix}$$</span></p>
<p>The problem inside the parentheses yields: <span class="math-container">$x_1(p,x_3,M)=x_2(p,x_3,M)=\frac{M-p_3x_3}{p_1+p_2}$</span>
Notice that the solution to the problem inside the brackets is similar to a demand function for <span class="math-container">$u=\min(x_1,x_2)$</span> with the difference being that we adjusted income in the demand function for the expenditure that might occur on <span class="math-container">$x_3$</span></p>
<p>Stage Two: Solving the entire problem in terms of <span class="math-container">$x_3$</span> after obtaining demands for <span class="math-container">$x_1,x_2$</span> as a function of <span class="math-container">$x_3$</span>.
Substituting the values of <span class="math-container">$x_1,x_2$</span> that we derived above in the problem inside the parentheses gives us the following optimization problem in <span class="math-container">$x_3$</span>:
<span class="math-container">$$\begin{aligned}
\max_{0\leq x_3\leq \frac{M}{p_3}} \quad &amp; x_3+\frac{M-p_3x_3}{p_1+p_2} \\
\max_{0\leq x_3\leq \frac{M}{p_3}} \quad &amp; \frac{(p_1+p_2-p_3)x_3}{p_1+p_2}+\frac{M}{p_1+p_2}
\end{aligned}$$</span></p>
<p>from the above problem, we get the demand function for <span class="math-container">$x_3$</span> and consequently the demand functions for <span class="math-container">$x_1, x_2$</span> by substituting the demand for <span class="math-container">$x_3$</span> in our solution to the Stage One problem.</p>
<p><span class="math-container">$\boxed{(x_1,x_2,x_3)^d(p,M)=\begin{cases}(\frac{M}{p_3},0,0) &amp; \text{if } p_1+p_2&gt;p_3\\
(\frac{M}{p_1+p_2},\frac{M}{p_1+p_2},0) &amp; \text{if } p_1+p_2\leq p_3\end{cases}}$</span></p>
<p>Intuitively it is like a solution to a perfect substitute utility function, but one of the goods is a composite good consisting of two goods which are complementary.</p>

CC BY-SA 4.0
2023-04-28T19:06:14.947Z
2023-04-28T19:17:44.767Z
2023-04-28T19:17:44.767Z
mynameparv
<p>I know how to solve the two-good case with <span class="math-container">$u(x) = \min\{x_1, x_2\}$</span>, but the addition of <span class="math-container">$x_3$</span> confuses me.</p>
<p><strong>Problem</strong></p>
<p>Derive the demand function <span class="math-container">$x(p,w)$</span> from <span class="math-container">$u(x) = \min\{x_1, x_2\} + x_3$</span>.</p>
<p><strong>What I did so far</strong></p>
<p>We assume that in optimum <span class="math-container">$x_1 = x_2$</span>.</p>
<p>Set up the budget constraint <span class="math-container">$p_1x_1 + p_2x_2 + p_3x_3 = w$</span>.</p>
<p>Rewrite budget constraint as <span class="math-container">$(p_1+p_2)x_1 + p_3x_3 = w$</span> or <span class="math-container">$(p_1+p_2)x_2 + p_3x_3 = w$</span>.</p>
<p>We can write <span class="math-container">$x_1^*=x_2^* = \frac{w-p_3x_3}{p_1+p_2}$</span> and <span class="math-container">$x_3^* = \frac{w-(p_1+p_2)x_1^*}{p_3}$</span>.</p>
<p><strong>Confusion</strong></p>
<p>How to proceed? Can I still use a Lagrangian to solve this?</p>

<microeconomics><consumer-theory><leontief><walrasian>
Derive demand function $x(p,w)$ from utility function $u(x) = \min\{x_1, x_2\} + x_3$
Anne1005
# Derive demand function $x(p,w)$ from utility function $u(x) = \min\{x_1, x_2\} + x_3$

(Asked by: username_0 on 2019-10-09)

I know how to solve the two-good case with $u(x) = \min\{x\_1, x\_2\}$, but the addition of $x\_3$ confuses me.

**Problem**

Derive the demand function $x(p,w)$ from $u(x) = \min\{x\_1, x\_2\} + x\_3$.

**What I did so far**

We assume that in optimum $x\_1 = x\_2$.

Set up the budget constraint $p\_1x\_1 + p\_2x\_2 + p\_3x\_3 = w$.

Rewrite budget constraint as $(p\_1+p\_2)x\_1 + p\_3x\_3 = w$ or $(p\_1+p\_2)x\_2 + p\_3x\_3 = w$.

We can write $x\_1^\*=x\_2^\* = \frac{w-p\_3x\_3}{p\_1+p\_2}$ and $x\_3^\* = \frac{w-(p\_1+p\_2)x\_1^\*}{p\_3}$.

**Confusion**

How to proceed? Can I still use a Lagrangian to solve this?

**Tags:** microeconomics, consumer-theory, leontief, walrasian


---

## 4 answers

### Answer by username_1 on 2022-11-14. Score: 1

We solve the problem

\begin{equation}
\max U(x\_1,x\_2,x\_3) = \min\{x\_1,x\_2\} + x\_3
\end{equation}

subject to

\begin{equation}
x\_1 p\_1 + x\_2 p\_2 + x\_3 p\_3 = I
\end{equation}

From the min term we get that $x\_1 = x\_2$. Therefore the budget constraint becomes

\begin{equation}
x\_1 (p\_1 + p\_2) + x\_3 p\_3 = I
\end{equation}

Solving for $x\_1$ we get

\begin{equation}
x\_1 = \frac{I - x\_3 p\_3}{p\_1 + p\_2}
\end{equation}

Since $x\_1 = x\_2$,

\begin{equation}
x\_2 = \frac{I - x\_3 p\_3}{p\_1 + p\_2}
\end{equation}

Therefore, the min term becomes

\begin{equation}
\min\{x\_1,x\_2\} = \frac{I - x\_3 p\_3}{p\_1 + p\_2}
\end{equation}

Substituting this expression for the min term into the utility function we get

\begin{equation}
W(x\_3) := U(x\_1(x\_3),x\_2(x\_3),x\_3) = \frac{I - x\_3 p\_3}{p\_1 + p\_2} + x\_3
\end{equation}

Separating the fraction and rearranging we get

\begin{equation}
W(x\_3) = (1 - \frac{p\_3}{p\_1 + p\_2}) x\_3 + \frac{I}{p\_1 + p\_2}
\end{equation}

Note that this is a straight line, with slope $1 - \frac{p\_3}{p\_1 + p\_2}$.

Notice $1 - \frac{p\_3}{p\_1 + p\_2} > 0 \iff 1 > \frac{p\_3}{p\_1 + p\_2} \iff p\_1 + p\_2 > p\_3$.

From here we get 3 cases:

[![enter image description here](https://i.sstatic.net/PEzjH.png)](https://i.sstatic.net/PEzjH.png)

* $p\_1 + p\_2 > p\_3 \rightarrow$ spend everything on $x\_3 \rightarrow x\_1 = 0, x\_2 = 0, x\_3 = \frac{I}{p\_3}$.
* $p\_1 + p\_2 < p\_3 \rightarrow$ don't consume $x\_3 \rightarrow x\_1 = \frac{I}{p\_1 + p\_2}, x\_2 = \frac{I}{p\_1 + p\_2}, x\_3 = 0$
* $p\_1 + p\_2 = p\_3 \rightarrow x\_3$ value indifferent $\rightarrow x\_1 = \frac{I - x\_3 p\_3}{p\_1 + p\_2}, x\_2 = \frac{I - x\_3 p\_3}{p\_1 + p\_2}, 0 \leq x\_3 \leq \frac{I}{p\_3}$, i.e. the optimal bundles form a line in 3-D space that looks like this:

[![enter image description here](https://i.sstatic.net/vuCBD.png)](https://i.sstatic.net/vuCBD.png)


---

### Answer by username_2 on 2019-11-10. Score: 1

for this problem you must consider two possible branches of the utility function:

$$u(\text{x})=x\_1+x\_3\ \ \text{if} \ \ x\_1<x\_2$$
$$u(\text{x})=x\_2+x\_3\ \ \text{if} \ \ x\_1>x\_2$$

The demand equations you solve from each one of these provide guidance on what the solution would be to your problem. However you must list them for each case.

Hope this helps


---

### Answer by username_3 on 2019-10-09. Score: 1

Hint: Imagine that there are two coffee bars, $A$ and $B$. There is only one type of coffee in the world. My preferences are such that I always want 1 unit of sugar with 1 unit of coffee; if I consume units of coffee and sugar in the ratio $1:1$, additional units of only one of the two don't give me extra utility.

At coffee bar $A$ they sell coffee and sugar separately. That is: there is one counter for coffee, where coffee is sold for a price of $p\_1$ per unit. There is another counter for sugar, where I can buy one unit of sugar for a price of $p\_2$.

At coffee bar $B$ however, they only sell one unit of coffee with one unit of sugar in a bundle, the price of which is equal to $p\_3$. They do not sell anything separately

At what prices do I prefer to buy my coffee with sugar at $A$, at what prices do I prefer to buy my coffee with sugar at $B$?

> **Comments:**
> **username_0**: If I understand you correctly, you buy at bar $A$ for $p1+p2$, and at bar $B$ for $p3$.  
> **username_3**: That is correct, but what I meant with the last sentence was: At what values of $p_1$, $p_2$, $p_3$ is it cheaper to buy my coffee with sugar at $A$?  
> **username_3**: I edited the answer to make it more clear, try to relate the little story above to the  problem you have to solve  
> **username_0**: Thank you. In that case it means that you will buy at $A$ if $p1 + p2$ < $p3$. If $p1 + p2$ > $p3$, you will buy at $B$. If they are equal, you are indifferent.  
> **username_3**: Exactly! And if you now combine this with the fact that your budget is equal to $w$ you have your demand function.  

---

### Answer by username_4 on 2023-04-28. Score: 0

Since you already know how to solve for the two goods case with $u(x)=\min(x\_1,x\_2)$ you can solve the optimization problem in a two-stage method without using Lagrangian.

UMP: $$\begin{aligned}
\max\_{x\_1,x\_2,x\_3\geq 0} \quad & \min(x\_1,x\_2)+x\_3\\
\textrm{s.t.} \quad & p\_1x\_1+p\_2x\_2+p\_3x\_3\leq M\end{aligned}$$

Stage one: Solving the problem with respect to $x\_1$ and $x\_2$ holding $x\_3$ constant
$$\max\_{x\_3}\begin{pmatrix}{\begin{aligned}\max\_{x\_1,x\_2} \quad & \min(x\_1,x\_2)+x\_3\\
\textrm{s.t. } \quad & p\_1x\_1+p\_2x\_2\leq M-p\_3x\_3\end{aligned}}\end{pmatrix}$$

The problem inside the parentheses yields: $x\_1(p,x\_3,M)=x\_2(p,x\_3,M)=\frac{M-p\_3x\_3}{p\_1+p\_2}$
Notice that the solution to the problem inside the brackets is similar to a demand function for $u=\min(x\_1,x\_2)$ with the difference being that we adjusted income in the demand function for the expenditure that might occur on $x\_3$

Stage Two: Solving the entire problem in terms of $x\_3$ after obtaining demands for $x\_1,x\_2$ as a function of $x\_3$.
Substituting the values of $x\_1,x\_2$ that we derived above in the problem inside the parentheses gives us the following optimization problem in $x\_3$:
$$\begin{aligned}
\max\_{0\leq x\_3\leq \frac{M}{p\_3}} \quad & x\_3+\frac{M-p\_3x\_3}{p\_1+p\_2} \\
\max\_{0\leq x\_3\leq \frac{M}{p\_3}} \quad & \frac{(p\_1+p\_2-p\_3)x\_3}{p\_1+p\_2}+\frac{M}{p\_1+p\_2}
\end{aligned}$$

from the above problem, we get the demand function for $x\_3$ and consequently the demand functions for $x\_1, x\_2$ by substituting the demand for $x\_3$ in our solution to the Stage One problem.

$\boxed{(x\_1,x\_2,x\_3)^d(p,M)=\begin{cases}(\frac{M}{p\_3},0,0) & \text{if } p\_1+p\_2>p\_3\\
(\frac{M}{p\_1+p\_2},\frac{M}{p\_1+p\_2},0) & \text{if } p\_1+p\_2\leq p\_3\end{cases}}$

Intuitively it is like a solution to a perfect substitute utility function, but one of the goods is a composite good consisting of two goods which are complementary.


---

