<p>So, I have not really worked out the maths behind solving this. I prefer to take a shortcut and just compute it numerically. Here is what I have. </p>

<p>These are the contour surfaces and the budget constraint. Notice how the solution will be at the "kinked" curve in the middle of each surface. 
<a href="https://i.sstatic.net/4fydl.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4fydl.png" alt="enter image description here"></a></p>

<p>The levels 1 and 2 are only for example. The third level, u = 3.92762 is related to the next part of my answer. Also, BC is the budget curve with income = 300. I have chosen <span class="math-container">$\alpha = 0.5$</span> and <span class="math-container">$P_1 = 2, P_2=3, P_3=1$</span>. </p>

<p>Now, using the parameters mentioned in the last paragraph, the maximum is found to be at <span class="math-container">$q_1 = 65.8896, q_2 = 39.1448, q_3 = 50.7863$</span> and the maximized utility is <span class="math-container">$u = 3.92762$</span>. I assumed that all goods are strictly consumed at positive amounts. </p>

<p>I am attaching my Mathematica code below, so that you can replicate and modify these results as you want</p>

<blockquote>
  <p>u = Min[a*Log[q1] + ((1 - a)*Log[q2]), Log[q3]]   </p>
  
  <p>BC = p1*q1 + p2*q2 + p3*q3<br>
    a = 0.5; p1 = 2; p2 = 3; p3 = 1;</p>
  
  <p>ContourPlot3D[{u == 1, u == 2, u == 3.92762, BC == 300}, {q1, 0,<br>
  100}, {q2, 0, 100} , {q3, 0, 100}, AxesLabel -> Automatic,<br>
  ContourStyle -> {Opacity[0.5], Opacity[0.5], Opacity[0.5],     Opacity
  [1]}, Mesh -> None, PlotLegends -> "Expressions"]</p>
  
  <p>NMaximize[{u, BC &lt;= 300, q1 > 0, q2 > 0, q3 > 0}, {q1, q2, q3}, Reals]</p>
</blockquote>

CC BY-SA 4.0
2018-10-29T02:46:21.587Z
2018-10-29T03:08:10.580Z
2018-10-29T03:08:10.580Z
erik
<p>Consider a utility function on the form <span class="math-container">$u(q_{1},q_{2},q_{3}) = min\{\alpha ln(q_{1}) + (1 - \alpha) ln(q_{2}), ln(q_{3})\}$</span></p>

<p>I know that optimal behaviour requires <span class="math-container">$\alpha ln (q_{1}) + (1 - \alpha) ln(q_{2}) = ln(q_{3})$</span></p>

<p>I tried substituting this into the budget constraint (after raising both sides to the power of <span class="math-container">$e$</span>)</p>

<p><span class="math-container">$p_{1}q_{1} + p_{2}q_{2} = y$</span></p>

<p>But I am unsure how to proceed in finding the Marshallian demand, since I have two variables <span class="math-container">$(q_{1}, q_{2})$</span> and only one equation. I have tried different transformations but my main problem is that it feels like I'm missing an equation.</p>

<p>Any help is greatly appreciated. </p>

<p>EDIT: I just re-read the question and I misspecified the budget constraint! The consumer only spends on <span class="math-container">$q_{1}$</span> and <span class="math-container">$q_{2}$</span>.</p>

CC BY-SA 4.0
2018-10-28T11:36:44.180Z
So since $\ln(0)$ is not defined but $\lim_{q_3 \to 0} \ln(q_3) = -\infty$ the task is to maximize w.r.t. $q_1,q_2$ the formula
$$
\min\{\alpha \ln(q_{1}) + (1 - \alpha) \ln(q_{2}), -\infty\} = -\infty?
$$
Giskard
CC BY-SA 4.0
2018-10-28T11:45:57.393Z
I believe we just have to assume some exogenous non-zero $q_{3}$. In the question I am given that $q_{1}$ and $q_{2}$ represent tea and coffee and $q_{3}$ represents sugar. I tried a transformation such that the UMP became min$\{\phi_{1} + \phi_{2}, \phi_{3}\}$ and thus at optimum $\phi_{1} + \phi_{2} = \phi_{3}$. Wouldn't this mean that the agent chooses the cheaper one of $\phi_{1}$ and $\phi_{2}$?
Gensys
<microeconomics><utility><leontief>
Marshallian demand with Leontif preferences
Gensys
# Marshallian demand with Leontif preferences

(Asked by: username_0 on 2018-10-27)

Consider a utility function on the form $u(q\_{1},q\_{2},q\_{3}) = min\{\alpha ln(q\_{1}) + (1 - \alpha) ln(q\_{2}), ln(q\_{3})\}$

I know that optimal behaviour requires $\alpha ln (q\_{1}) + (1 - \alpha) ln(q\_{2}) = ln(q\_{3})$

I tried substituting this into the budget constraint (after raising both sides to the power of $e$)

$p\_{1}q\_{1} + p\_{2}q\_{2} = y$

But I am unsure how to proceed in finding the Marshallian demand, since I have two variables $(q\_{1}, q\_{2})$ and only one equation. I have tried different transformations but my main problem is that it feels like I'm missing an equation.

Any help is greatly appreciated.

EDIT: I just re-read the question and I misspecified the budget constraint! The consumer only spends on $q\_{1}$ and $q\_{2}$.

**Tags:** microeconomics, utility, leontief

> **Comments:**
> **username_1**: So since $\ln(0)$ is not defined but $\lim_{q_3 \to 0} \ln(q_3) = -\infty$ the task is to maximize w.r.t. $q_1,q_2$ the formula $$ \min\{\alpha \ln(q_{1}) + (1 - \alpha) \ln(q_{2}), -\infty\} = -\infty? $$  
> **username_0**: I believe we just have to assume some exogenous non-zero $q_{3}$. In the question I am given that $q_{1}$ and $q_{2}$ represent tea and coffee and $q_{3}$ represents sugar. I tried a transformation such that the UMP became min$\{\phi_{1} + \phi_{2}, \phi_{3}\}$ and thus at optimum $\phi_{1} + \phi_{2} = \phi_{3}$. Wouldn't this mean that the agent chooses the cheaper one of $\phi_{1}$ and $\phi_{2}$?  

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## 1 answers

### Answer by username_2 on 2018-10-29. Score: 1

So, I have not really worked out the maths behind solving this. I prefer to take a shortcut and just compute it numerically. Here is what I have.

These are the contour surfaces and the budget constraint. Notice how the solution will be at the "kinked" curve in the middle of each surface.
[![enter image description here](https://i.sstatic.net/4fydl.png)](https://i.sstatic.net/4fydl.png)

The levels 1 and 2 are only for example. The third level, u = 3.92762 is related to the next part of my answer. Also, BC is the budget curve with income = 300. I have chosen $\alpha = 0.5$ and $P\_1 = 2, P\_2=3, P\_3=1$.

Now, using the parameters mentioned in the last paragraph, the maximum is found to be at $q\_1 = 65.8896, q\_2 = 39.1448, q\_3 = 50.7863$ and the maximized utility is $u = 3.92762$. I assumed that all goods are strictly consumed at positive amounts.

I am attaching my Mathematica code below, so that you can replicate and modify these results as you want

> u = Min[a\*Log[q1] + ((1 - a)\*Log[q2]), Log[q3]]
>
> BC = p1\*q1 + p2\*q2 + p3\*q3  
> a = 0.5; p1 = 2; p2 = 3; p3 = 1;
>
> ContourPlot3D[{u == 1, u == 2, u == 3.92762, BC == 300}, {q1, 0,  
> 100}, {q2, 0, 100} , {q3, 0, 100}, AxesLabel -> Automatic,  
> ContourStyle -> {Opacity[0.5], Opacity[0.5], Opacity[0.5], Opacity
> [1]}, Mesh -> None, PlotLegends -> "Expressions"]
>
> NMaximize[{u, BC <= 300, q1 > 0, q2 > 0, q3 > 0}, {q1, q2, q3}, Reals]


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