<p>If we could get the demand function <span class="math-container">$c(p) = 1 - p$</span> (for <span class="math-container">$p\leq1$</span>) that would be great, because then
<span class="math-container">$$
\epsilon(p) = \frac{\text{d}c(p)}{\text{d}p}\frac{p}{c(p)} = -\frac{p}{1-p}
$$</span>
would run the gamut:
<span class="math-container">\begin{align*}
\lim_{p \searrow 0} \epsilon(p) &amp; = \lim_{p \searrow 0}  \frac{-p}{1-p} = 0 \\
\\
\lim_{p \nearrow 1} \epsilon(p) &amp; = \lim_{p \nearrow 1}  \frac{-p}{1-p} = -\infty,
\end{align*}</span>
with demand being inelastic under <span class="math-container">$p=1/2$</span>.</p>
<p>Good news! It is in fact possible to get this demand function from the utility function
<span class="math-container">$$
U(c,y) = c - \frac{c^2}{2} + y,
$$</span>
where <span class="math-container">$y$</span> is the amount of money spent on other goods.</p>
<hr />
<p>Could we get a demand function <span class="math-container">$c(p)$</span> where <span class="math-container">$\epsilon(p) = p^{\alpha}$</span>, with <span class="math-container">$0 &gt; \alpha &gt; -1$</span> over the entire price range? Bad news :( <br>
Such a demand function would imply an infinite budget, since the expenditure on the good is
<span class="math-container">$$
c(p) \cdot p = p^{\alpha} \cdot p = p^{\alpha + 1}, 
$$</span>
and since <span class="math-container">$ \alpha &gt; -1$</span>, we have
<span class="math-container">$$
\lim_{p \to \infty} c(p) \cdot p = \lim_{p \to \infty} p^{\alpha + 1} = \infty.
$$</span></p>

CC BY-SA 4.0
2023-12-19T08:02:28.373Z
Excellent, thank you! Does the utility function $U(c) = c - c^2/2$ have a name?
Anthony
CC BY-SA 4.0
2023-12-19T08:48:12.860Z
Hi! $U(c,y) = c - c^2/2 + y$ is *a* quasilinear utility function. I am unclear on what model you are using; $U(c) = c - c^2/2$ is a univariate utility function, it is not very meaningful in itself, without the model. You will get a different demand function if you are doing intertemporal decisions with this type of function.
Giskard
CC BY-SA 4.0
2023-12-19T09:23:06.333Z
Thanks, I am planning to use it in a stochastic control context - I was just looking for alternatives to power utility. Happy to give this one a whirl. Apparently this type of quadratic utility shows up everywhere in portfolio theory.
Anthony
CC BY-SA 4.0
2023-12-19T09:36:02.623Z
How would you modify your approach above for intertemporal decisions?
Anthony
CC BY-SA 4.0
2023-12-19T09:53:06.647Z
Hi! This is a very far leading and separate question. Please post it as a new question and give enough details to make an answer possible.
Giskard
CC BY-SA 4.0
2023-12-18T23:34:50.167Z
2023-12-18T23:43:20.867Z
2023-12-18T23:43:20.867Z
Giskard
<p>I am looking for a example of a utility function which, when the utility maximisation problem is solved, results in an inelastic demand. The standard examples in textbooks always seem to have unitary elasticity.</p>
<p>In more detail, is it possible to find an example of a utility function <span class="math-container">$U(c)$</span> such that solving <span class="math-container">$\sup_c U(c)$</span> subject to <span class="math-container">$p c \leq m$</span> results in a consumption rule <span class="math-container">$c^*(p, m) \propto p^{-\alpha}$</span>, with <span class="math-container">$\alpha &lt; 1$</span>?</p>
<p>References are always appreciated.</p>

<reference-request><utility><consumer-theory><elasticity>
Example of a utility function which yields inelastic demand function
Anthony
# Example of a utility function which yields inelastic demand function

(Asked by: username_0 on 2023-12-18)

I am looking for a example of a utility function which, when the utility maximisation problem is solved, results in an inelastic demand. The standard examples in textbooks always seem to have unitary elasticity.

In more detail, is it possible to find an example of a utility function $U(c)$ such that solving $\sup\_c U(c)$ subject to $p c \leq m$ results in a consumption rule $c^\*(p, m) \propto p^{-\alpha}$, with $\alpha < 1$?

References are always appreciated.

**Tags:** reference-request, utility, consumer-theory, elasticity

---

## 1 answers

### Answer by username_1 on 2023-12-18. Score: 2 | Accepted answer

If we could get the demand function $c(p) = 1 - p$ (for $p\leq1$) that would be great, because then
$$
\epsilon(p) = \frac{\text{d}c(p)}{\text{d}p}\frac{p}{c(p)} = -\frac{p}{1-p}
$$
would run the gamut:
\begin{align\*}
\lim\_{p \searrow 0} \epsilon(p) & = \lim\_{p \searrow 0} \frac{-p}{1-p} = 0 \\
\\
\lim\_{p \nearrow 1} \epsilon(p) & = \lim\_{p \nearrow 1} \frac{-p}{1-p} = -\infty,
\end{align\*}
with demand being inelastic under $p=1/2$.

Good news! It is in fact possible to get this demand function from the utility function
$$
U(c,y) = c - \frac{c^2}{2} + y,
$$
where $y$ is the amount of money spent on other goods.

---

Could we get a demand function $c(p)$ where $\epsilon(p) = p^{\alpha}$, with $0 > \alpha > -1$ over the entire price range? Bad news :(   
Such a demand function would imply an infinite budget, since the expenditure on the good is
$$
c(p) \cdot p = p^{\alpha} \cdot p = p^{\alpha + 1},
$$
and since $ \alpha > -1$, we have
$$
\lim\_{p \to \infty} c(p) \cdot p = \lim\_{p \to \infty} p^{\alpha + 1} = \infty.
$$

> **Comments:**
> **username_0**: Excellent, thank you! Does the utility function $U(c) = c - c^2/2$ have a name?  
> **username_1**: Hi! $U(c,y) = c - c^2/2 + y$ is *a* quasilinear utility function. I am unclear on what model you are using; $U(c) = c - c^2/2$ is a univariate utility function, it is not very meaningful in itself, without the model. You will get a different demand function if you are doing intertemporal decisions with this type of function.  
> **username_0**: Thanks, I am planning to use it in a stochastic control context - I was just looking for alternatives to power utility. Happy to give this one a whirl. Apparently this type of quadratic utility shows up everywhere in portfolio theory.  
> **username_0**: How would you modify your approach above for intertemporal decisions?  
> **username_1**: Hi! This is a very far leading and separate question. Please post it as a new question and give enough details to make an answer possible.  

---

