<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
2022-11-14T02:13:54.370Z
2022-11-14T02:13:54.370Z
Nicolas Torres
<p>I will give a hint. To increase/decrease utility by 1 unit, you can do two things. Increase $x_1$ and $x_2$ by $\frac{1}{2}$ units and this would cost you $\frac{p_1+p_2}{2}$. The second option is changing $x_3$ by one unit that would cost $p_3$. One other option is of course using a linear combination of the above changes. If you remember from well behaved utility functions, you would like to buy things for which utility per price is higher. Now you would go for a similar reasoning here and would try to buy extra utility at a cheapest price.</p>

CC BY-SA 3.0
2015-09-30T00:17:31.220Z
I think i get it. Thank you, if something happens I shall ask again.
Marcos Fabian C
CC BY-SA 3.0
2015-09-30T00:05:30.870Z
2015-09-30T00:05:30.870Z
ramazan
<p>Let an agent with utility function $U(x_1,x_2,x_3)=min\{ x_1, x_2 \} +x_3$. We want to maximize $U$ subject to $p_1 x_1+p_2 x_2 +p_3 x_3= I$</p>

<p>First intuitively $x_1^*=x_2^*$ substituting en the budget constraint I have $$x_1^*=x_2^*=\frac{I-p_3 x_3}{p_1+p_2}$$ I don't know how to solve for $x_3$. Then i thought in Kuhn-Tucker but I can't see how to solve it.</p>

<microeconomics>
About marshallian demands
Marcos Fabian C
# About marshallian demands

(Asked by: username_0 on 2015-09-29)

Let an agent with utility function $U(x\_1,x\_2,x\_3)=min\{ x\_1, x\_2 \} +x\_3$. We want to maximize $U$ subject to $p\_1 x\_1+p\_2 x\_2 +p\_3 x\_3= I$

First intuitively $x\_1^\*=x\_2^\*$ substituting en the budget constraint I have $$x\_1^\*=x\_2^\*=\frac{I-p\_3 x\_3}{p\_1+p\_2}$$ I don't know how to solve for $x\_3$. Then i thought in Kuhn-Tucker but I can't see how to solve it.

**Tags:** microeconomics


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## 2 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)


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### Answer by username_2 on 2015-09-30. Score: 1

I will give a hint. To increase/decrease utility by 1 unit, you can do two things. Increase $x\_1$ and $x\_2$ by $\frac{1}{2}$ units and this would cost you $\frac{p\_1+p\_2}{2}$. The second option is changing $x\_3$ by one unit that would cost $p\_3$. One other option is of course using a linear combination of the above changes. If you remember from well behaved utility functions, you would like to buy things for which utility per price is higher. Now you would go for a similar reasoning here and would try to buy extra utility at a cheapest price.

> **Comments:**
> **username_0**: I think i get it. Thank you, if something happens I shall ask again.  

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