<p>Given the utility maximisation problem,</p>
<p><span class="math-container">\begin{eqnarray*}\max_{(x_1,x_2,x_3)\in\mathbb{R}^3_+} &amp; \min\left(2x_1+x_3,\frac{x_2}{2}\right) \\ \text{s.t. } &amp; p_1x_1+p_2x_2+p_3x_3\leq M\end{eqnarray*}</span>
where <span class="math-container">$p_1&gt;0$</span>, <span class="math-container">$p_2&gt;0$</span>, <span class="math-container">$p_3&gt;0$</span>, and <span class="math-container">$M\geq 0$</span> are given.</p>
<p>Solving this problem, we get the Marshallian demand as
<span class="math-container">\begin{eqnarray*}(x_1^d,x_2^d,x_3^d)(p_1,p_2,p_3,M)\in \begin{cases} \left\{\left(\dfrac{M}{p_1+4p_2},\dfrac{4M}{p_1+4p_2},0\right)\right\} &amp; \text{if } p_1&lt;2p_3 \\ \left\{\left(0,\dfrac{2M}{2p_2+p_3},\dfrac{M}{2p_2+p_3}\right)\right\} &amp; \text{if } p_1&gt;2p_3 \\ \left\{\left(\dfrac{\theta M}{p_1+4p_2},\dfrac{4M}{p_1+4p_2},\dfrac{(1-\theta)M}{2p_2+p_3}\right)|\theta\in[0,1]\right\} &amp; \text{if } p_1=2p_3\end{cases}\end{eqnarray*}</span></p>

CC BY-SA 4.0
2024-08-20T20:07:34.580Z
Why are we comparing p1 with 2p3 instead of just p3? And why are we comparing to P3 and not P2, my solution manual uses p2/2
AnonGal
CC BY-SA 4.0
2024-08-21T08:11:19.837Z
Because we need to choose between $x_1$ and $x_3$ since the two are substitutes, and here the expression is $2x_1+x_3$.
Amit
CC BY-SA 4.0
2024-08-20T18:27:00.727Z
2024-08-20T18:27:00.727Z
Amit
<p>I have a utility function</p>
<p><span class="math-container">$$u(x_1, x_2, x_3) = \min \{2x_1 + x_3, x_2/2\} $$</span></p>
<p>I would have assumed that the relationship established is <span class="math-container">$2x_1 + x_3 = x_2/2$</span> but my solution manual has it as  <span class="math-container">$$x_1 + 2x_2 = x_3/2 $$</span></p>
<p>I'm very confused</p>

CC BY-SA 4.0
2024-08-20T17:59:52.643Z
You are right, the optimal choice i.e. the Marshallian demand in this problem satisfy $2x_1+x_3=\frac{x_2}{2}$.
Amit
CC BY-SA 4.0
2024-08-20T18:04:15.900Z
Your solution manual appears to be wrong, unless there's a typo or the suggested solution is a simplification after factoring in the prices (which have not been provided here).
Herr K.
CC BY-SA 4.0
2024-08-20T19:50:05.543Z
Yeah I was worried about that. My workout reaches as far as rearranching the relationship and substituting x2 = 4x1 + 2x3 into the endowment function/ ("budget constraint")  to obtain m = p1x2 + 4p2x1 + 2p2x3 + p3x3 but I'm not sure how to progress from that point
AnonGal
<microeconomics><optimization>
How to find marshallian demand for Leontief Utility with 3 goods. u(x 1 ​ ,x 2 ​ ,x 3 ​)=min{2x1 + x3, x2/2}
AnonGal
# How to find marshallian demand for Leontief Utility with 3 goods. u(x 1 ​ ,x 2 ​ ,x 3 ​)=min{2x1 + x3, x2/2}

(Asked by: username_0 on 2024-08-20)

I have a utility function

$$u(x\_1, x\_2, x\_3) = \min \{2x\_1 + x\_3, x\_2/2\} $$

I would have assumed that the relationship established is $2x\_1 + x\_3 = x\_2/2$ but my solution manual has it as $$x\_1 + 2x\_2 = x\_3/2 $$

I'm very confused

**Tags:** microeconomics, optimization
> **Comments:**
> **username_1**: You are right, the optimal choice i.e. the Marshallian demand in this problem satisfy $2x_1+x_3=\frac{x_2}{2}$.  
> **username_2**: Your solution manual appears to be wrong, unless there's a typo or the suggested solution is a simplification after factoring in the prices (which have not been provided here).  
> **username_0**: Yeah I was worried about that. My workout reaches as far as rearranching the relationship and substituting x2 = 4x1 + 2x3 into the endowment function/ ("budget constraint")  to obtain m = p1x2 + 4p2x1 + 2p2x3 + p3x3 but I'm not sure how to progress from that point  

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## 1 answers

### Answer by username_1 on 2024-08-20. Score: 0

Given the utility maximisation problem,

\begin{eqnarray\*}\max\_{(x\_1,x\_2,x\_3)\in\mathbb{R}^3\_+} & \min\left(2x\_1+x\_3,\frac{x\_2}{2}\right) \\ \text{s.t. } & p\_1x\_1+p\_2x\_2+p\_3x\_3\leq M\end{eqnarray\*}
where $p\_1>0$, $p\_2>0$, $p\_3>0$, and $M\geq 0$ are given.

Solving this problem, we get the Marshallian demand as
\begin{eqnarray\*}(x\_1^d,x\_2^d,x\_3^d)(p\_1,p\_2,p\_3,M)\in \begin{cases} \left\{\left(\dfrac{M}{p\_1+4p\_2},\dfrac{4M}{p\_1+4p\_2},0\right)\right\} & \text{if } p\_1<2p\_3 \\ \left\{\left(0,\dfrac{2M}{2p\_2+p\_3},\dfrac{M}{2p\_2+p\_3}\right)\right\} & \text{if } p\_1>2p\_3 \\ \left\{\left(\dfrac{\theta M}{p\_1+4p\_2},\dfrac{4M}{p\_1+4p\_2},\dfrac{(1-\theta)M}{2p\_2+p\_3}\right)|\theta\in[0,1]\right\} & \text{if } p\_1=2p\_3\end{cases}\end{eqnarray\*}

> **Comments:**
> **username_0**: Why are we comparing p1 with 2p3 instead of just p3? And why are we comparing to P3 and not P2, my solution manual uses p2/2  
> **username_1**: Because we need to choose between $x_1$ and $x_3$ since the two are substitutes, and here the expression is $2x_1+x_3$.  

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