<p>Let <span class="math-container">$K:=y-(f+fa)-(t+ta)x$</span>, and so <span class="math-container">$c=K-pq$</span>.</p>
<p>From condition <span class="math-container">$(\mathrm{A1})$</span>,
<span class="math-container">\begin{align}
&amp;&amp;\frac{v_q(\cdot)}{v_c(\cdot)}&amp;=p \\
\quad\Rightarrow\quad &amp;&amp;
\frac{\alpha[K-pq]^{1-\alpha}q^{\alpha-1}}{(1-\alpha)[K-pq]^{-\alpha}q^\alpha}&amp;=p \\
\quad\Rightarrow\quad &amp;&amp;
pq&amp;=\alpha K \\
\quad\Rightarrow\quad &amp;&amp;
q&amp;=\frac{\alpha K}{p}. \tag{*}
\end{align}</span>
From <span class="math-container">$(\mathrm{A2})$</span> and <span class="math-container">$(*)$</span>,
<span class="math-container">\begin{align}
&amp;&amp;[K-pq]^{1-\alpha}q^\alpha &amp;= u \\
\quad\overset{(*)}{\Rightarrow}\quad 
&amp;&amp;[(1-\alpha)K]^{1-\alpha}\biggl(\frac{\alpha K}{p}\biggr)^\alpha &amp;= u\\
\quad\Rightarrow\quad
&amp;&amp; (1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}&amp;=p^\alpha. 
\end{align}</span>
Thus,
<span class="math-container">\begin{align}
p(\cdot)&amp;=\left[(1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}\right]^{1/\alpha}\\
q(\cdot)&amp;=\frac{\alpha K}{\left[(1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}\right]^{1/\alpha}}
\end{align}</span></p>

CC BY-SA 4.0
2022-01-27T08:55:11.530Z
So in A2 you first substitute in the c from the constraint and then you substitute in all q's with the marshallian demand function. And then just solve for p and q respectively.
Victor Nielsen
CC BY-SA 4.0
2022-01-27T19:17:16.527Z
I've been trying to do this myself with the other utility function. c^a * q^(1-a) because it seems the programme was written with that despite their paper writing the other one, but I don't get the desired result either using that. Instead, I get p = -(a-1)K(u(aK)^(-a))^(1/1-a). Wondering what I'm getting wrong here.
Victor Nielsen
CC BY-SA 4.0
2022-01-27T22:16:38.703Z
@VictorNielsen: If utility is $c^\alpha q^{1-\alpha}$, then the only difference in $p(\cdot)$ is the outer-most exponent. You probably made some algebraic mistakes somewhere, and without seeing the steps, it's hard to tell exactly where you got tripped up.
Herr K.
CC BY-SA 4.0
2022-01-28T10:52:07.660Z
if I switch around the q and c, then I get that the q in A1 is (K-aK)/p.
But if the equation's only difference is the exponent, how com your p(⋅) and q(⋅) be so different from the equations in the first picture in my post?
Victor Nielsen
CC BY-SA 4.0
2022-01-28T14:14:15.880Z
@VictorNielsen: $(K-\alpha K)/p=(1-\alpha)K/p$, which is similar to my $\alpha K/p$, except that $\alpha$ switches place with $(1-\alpha)$. And I don't think my results are very different from the coded expression in your post: switch $\alpha$ and $(1-\alpha)$ and plug in for $K$, you'll see they are the same.
Herr K.
CC BY-SA 4.0
2022-01-28T17:18:35.047Z
I see it now. Seems pretty obvious now. I can tell it's been 3 years since I worked with equations. Thanks for all the help!
Victor Nielsen
CC BY-SA 4.0
2022-01-27T00:57:17.417Z
2022-01-27T00:57:17.417Z
Herr K.
<p>I need to get this resulting price and quantity (housing):</p>
<p><a href="https://i.sstatic.net/P44iO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/P44iO.png" alt="price and quantity functions" /></a></p>
<p>It's pretty clear that the denominator of the quantity function is just the price function.</p>
<p>From this utility function:</p>
<p><a href="https://i.sstatic.net/34jWr.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/34jWr.png" alt="utility" /></a></p>
<p>And this constraint:</p>
<p><a href="https://i.sstatic.net/UrseY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UrseY.png" alt="constraint" /></a></p>
<p>Now this is a monocentric city model function, so these methods follow:</p>
<p><a href="https://i.sstatic.net/vOwBe.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vOwBe.png" alt="enter image description here" /></a></p>
<p>or:</p>
<p><a href="https://i.sstatic.net/QTPjf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QTPjf.png" alt="enter image description here" /></a></p>
<p>However, I do not see the actual process of doing that. I need to make this calculations for other utility functions, but when I follow the steps, I do not get the desired result.</p>

CC BY-SA 4.0
2022-01-26T20:48:19.630Z
Are you sure the "results" at the beginning are correct? They seem to correspond to the utility function $c^\alpha q^{1-\alpha}$, while your utility function is $c^{1-\alpha}q^\alpha$.
Herr K.
CC BY-SA 4.0
2022-01-26T21:36:17.090Z
I'm not sure until I can calculate the steps. I'm using a programme developed by someone else, so I'm trying to trace back the steps. If that is the case, I will change the "result" to match the correct utility function.
Victor Nielsen
CC BY-SA 4.0
2022-01-27T17:42:35.763Z
@Victor 'I'm using a programme developed by someone else' is that program publicly available?
Jesper Hybel
CC BY-SA 4.0
2022-01-28T10:48:02.193Z
The programme is mathematica, but the script is made by a journal article, where you can ask for it.
Victor Nielsen
<microeconomics><utility><optimization><price><demand>
How do I get to this demand function in the monocentric city model?
Victor Nielsen
# How do I get to this demand function in the monocentric city model?

(Asked by: username_0 on 2022-01-26)

I need to get this resulting price and quantity (housing):

[![price and quantity functions](https://i.sstatic.net/P44iO.png)](https://i.sstatic.net/P44iO.png)

It's pretty clear that the denominator of the quantity function is just the price function.

From this utility function:

[![utility](https://i.sstatic.net/34jWr.png)](https://i.sstatic.net/34jWr.png)

And this constraint:

[![constraint](https://i.sstatic.net/UrseY.png)](https://i.sstatic.net/UrseY.png)

Now this is a monocentric city model function, so these methods follow:

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

or:

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

However, I do not see the actual process of doing that. I need to make this calculations for other utility functions, but when I follow the steps, I do not get the desired result.

**Tags:** microeconomics, utility, optimization, price, demand
> **Comments:**
> **username_1**: Are you sure the "results" at the beginning are correct? They seem to correspond to the utility function $c^\alpha q^{1-\alpha}$, while your utility function is $c^{1-\alpha}q^\alpha$.  
> **username_0**: I'm not sure until I can calculate the steps. I'm using a programme developed by someone else, so I'm trying to trace back the steps. If that is the case, I will change the "result" to match the correct utility function.  
> **username_2**: @Victor 'I'm using a programme developed by someone else' is that program publicly available?  
> **username_0**: The programme is mathematica, but the script is made by a journal article, where you can ask for it.  

---

## 1 answers

### Answer by username_1 on 2022-01-27. Score: 4 | Accepted answer

Let $K:=y-(f+fa)-(t+ta)x$, and so $c=K-pq$.

From condition $(\mathrm{A1})$,
\begin{align}
&&\frac{v\_q(\cdot)}{v\_c(\cdot)}&=p \\
\quad\Rightarrow\quad &&
\frac{\alpha[K-pq]^{1-\alpha}q^{\alpha-1}}{(1-\alpha)[K-pq]^{-\alpha}q^\alpha}&=p \\
\quad\Rightarrow\quad &&
pq&=\alpha K \\
\quad\Rightarrow\quad &&
q&=\frac{\alpha K}{p}. \tag{\*}
\end{align}
From $(\mathrm{A2})$ and $(\*)$,
\begin{align}
&&[K-pq]^{1-\alpha}q^\alpha &= u \\
\quad\overset{(\*)}{\Rightarrow}\quad
&&[(1-\alpha)K]^{1-\alpha}\biggl(\frac{\alpha K}{p}\biggr)^\alpha &= u\\
\quad\Rightarrow\quad
&& (1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}&=p^\alpha.
\end{align}
Thus,
\begin{align}
p(\cdot)&=\left[(1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}\right]^{1/\alpha}\\
q(\cdot)&=\frac{\alpha K}{\left[(1-\alpha)^{1-\alpha}\alpha^\alpha \frac{K}{u}\right]^{1/\alpha}}
\end{align}

> **Comments:**
> **username_0**: So in A2 you first substitute in the c from the constraint and then you substitute in all q's with the marshallian demand function. And then just solve for p and q respectively.  
> **username_0**: I've been trying to do this myself with the other utility function. c^a * q^(1-a) because it seems the programme was written with that despite their paper writing the other one, but I don't get the desired result either using that. Instead, I get p = -(a-1)K(u(aK)^(-a))^(1/1-a). Wondering what I'm getting wrong here.  
> **username_1**: @VictorNielsen: If utility is $c^\alpha q^{1-\alpha}$, then the only difference in $p(\cdot)$ is the outer-most exponent. You probably made some algebraic mistakes somewhere, and without seeing the steps, it's hard to tell exactly where you got tripped up.  
> **username_0**: if I switch around the q and c, then I get that the q in A1 is (K-aK)/p. But if the equation's only difference is the exponent, how com your p(⋅) and q(⋅) be so different from the equations in the first picture in my post?  
> **username_1**: @VictorNielsen: $(K-\alpha K)/p=(1-\alpha)K/p$, which is similar to my $\alpha K/p$, except that $\alpha$ switches place with $(1-\alpha)$. And I don't think my results are very different from the coded expression in your post: switch $\alpha$ and $(1-\alpha)$ and plug in for $K$, you'll see they are the same.  
> **username_0**: I see it now. Seems pretty obvious now. I can tell it's been 3 years since I worked with equations. Thanks for all the help!  

---

