<p>As your cost function exhibits non constant first and second order derivative, a third degree polynomial is a good starting point.</p>
<p>There are infinitely many production functions behind a cubic cost function. Below there is just one example, using a trick from Gorman, which consists to interpret a cost function whose expression is the sum of three (or four) parts, as if it was obtained from a single firm using three machines (or three plants).<br />
Machine (or plant) number <span class="math-container">$j=1,...,3$</span>, produces a constant share <span class="math-container">$s_j$</span> of total output <span class="math-container">$y$</span> according to its own production function <span class="math-container">$h_j$</span> such that <span class="math-container">$$s_j y = h_j(x_j-k_j),$$</span> where each <span class="math-container">$h_j$</span> is homogeneous of degree <span class="math-container">$1/\alpha_j$</span> in <span class="math-container">$(x_j-k_j)$</span>, and <span class="math-container">$k_j \geq 0$</span> denotes the minimum input requirement for production: <span class="math-container">$h_j(x_j-k_j)=0$</span> for <span class="math-container">$x_j \leq k_j$</span>. Output is denoted by <span class="math-container">$y$</span> and the input price vector by <span class="math-container">$w$</span> (it can be considered as constant here).
Then the optimal input demand vector takes the form (left as exercise):
<span class="math-container">$$x_j^*(w,y)=k_j + b_j(w)y^{\alpha_j}$$</span>
and the corresponding cost function is
<span class="math-container">$$c(w,y) = \sum_{j=1}^3 w^T x_j^*(w,y) = a_0(w) + a_1(w)y^{\alpha_1} + a_2(w)y^{\alpha_2} + a_3(w)y^{\alpha_3}$$</span>
which is the expression of the cubic cost function for <span class="math-container">$\alpha_1=1,\alpha_2=2,$</span> and <span class="math-container">$\alpha_3=3$</span>.<br />
Further restrictions on the <span class="math-container">$a_j$</span> function yield the above figures.</p>

CC BY-SA 4.0
2021-11-02T09:33:35.293Z
I guess $k_j$ is the minimum amount of input $j$ required to produce anything? More generally, it might be good to define all the symbols in this answer (or at least provide a link).
afreelunch
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2021-11-02T16:40:51.603Z
@afreelunch: yes it is, I have added this point in the answer. As welle as some further explanations.
Bertrand
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2021-11-02T20:27:17.253Z
Thanks, that now makes more sense! Final question: can I just think of the $b_j(w)$ and $a_i(w)$ terms as positive constants?
afreelunch
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2021-11-02T21:40:52.180Z
Indeed, the functions $a_j$ and $b_j$ are constant in $y$ and are related by $a_j(w)=w^Tb_j(w)$ for $j=1,2,3$.
Bertrand
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2021-11-01T18:48:21.097Z
2021-11-02T16:38:56.790Z
2021-11-02T16:38:56.790Z
Bertrand
<p>As I noted in a comment, one approach is to allow for fixed costs (unlike in the graph). For example, suppose that production requires two factors <span class="math-container">$L$</span> and <span class="math-container">$K$</span> and that the function is symmetric Cobb-Douglas:
<span class="math-container">$$
q = L^\alpha K^\alpha \text{ where } \alpha \in (0, 1)
$$</span>
Solving the firm's cost minimisation problem then <a href="http://www.econ.ucla.edu/sboard/teaching/econ11_09/econ11_09_slides7.pdf" rel="nofollow noreferrer">yields</a> the cost function
<span class="math-container">$$
C(q) = 2\sqrt{wr} q^{\frac{1}{2\alpha}} \equiv c q^{\frac{1}{2\alpha}}
$$</span>
where <span class="math-container">$w$</span> and <span class="math-container">$r$</span> are the prices of <span class="math-container">$L$</span> and <span class="math-container">$K$</span> and <span class="math-container">$c&gt;0$</span> is a constant that we've just defined. If we assume you also need to pay a fixed cost <span class="math-container">$F &gt; 0$</span> to produce anything at all, then
<span class="math-container">$$
C(q)=
\begin{cases}
c q^{\frac{1}{2\alpha}} + F \text{ if } q &gt; 0\\
0 \hspace{3em}\text{ if $q = 0$ }
\end{cases}
$$</span>
If you choose <span class="math-container">$\alpha &lt; 0.5$</span> (decreasing returns to scale), then you get a convex cost function with a discontinuity at zero. For example, if <span class="math-container">$\alpha = 0.25$</span>, then costs start at <span class="math-container">$F$</span>, fall discontinuously, and then increase quadratically in <span class="math-container">$q$</span>. As a result, marginal costs are 'infinite' for the first unit, then drop discontinuously, and then rise as desired. Finally, average costs decrease and then rise. I realise that this isn't quite what you are after -- but it seems close.</p>

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2021-11-01T13:40:58.787Z
2021-11-02T20:48:09.363Z
2021-11-02T20:48:09.363Z
afreelunch
<p>I would like a production function that gives a cost function with the following shape:</p>
<p><a href="https://i.sstatic.net/Yv984.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Yv984.png" alt="enter image description here" /></a></p>
<p>The figure was taken from &quot;Microeconomic Theory: Basic Principles and Extensions, 12th edition&quot;, on Chapter 10, section 10.4.3</p>
<p>The cost function <span class="math-container">$c$</span> should satisfy the following requirements:</p>
<ol>
<li><span class="math-container">$c(q)$</span> is defined and continuous for all <span class="math-container">$q \geq 0$</span></li>
<li><span class="math-container">$c(q)$</span> is twice continuously differentiable for all <span class="math-container">$q &gt; 0$</span></li>
<li><span class="math-container">$c(q) \to \infty$</span> as <span class="math-container">$q \to \infty$</span></li>
<li>For all <span class="math-container">$q &gt; 0$</span>, we have <span class="math-container">$c'(q) &gt; 0$</span></li>
<li>There exists <span class="math-container">$q_1 &gt; 0$</span> such that <span class="math-container">$c''(q) &lt; 0$</span> for all <span class="math-container">$0 &lt; q &lt; q_1$</span> and <span class="math-container">$c''(q) &gt; 0$</span> for all <span class="math-container">$q &gt; q_1$</span></li>
</ol>
<p>I don't need <span class="math-container">$c$</span> to be literally a cubic polynomial (i.e. <span class="math-container">$ c(q) = a_0 + a_1 q + a_2 q^2 + a_3 q^3 $</span>). I just need it to look like the shape in the figure and satisfy the above requirements.</p>

CC BY-SA 4.0
2021-11-01T10:19:23.377Z
Do you require that $c(q) \rightarrow 0$ as $q \rightarrow 0$ (as in the picture)? If not, should be easy to do this using fixed costs. Also, do you want more than one factor input? It's quite trivial if you just have one input.
afreelunch
CC BY-SA 4.0
2021-11-01T11:16:46.943Z
@afreelunch Thank you for your comment. I don't need $c(0) = 0$ because as you said, it can be done with fixed costs. And yes, I want more then one input, otherwise it would be trivial.
user141240
<microeconomics><production-function>
What kind of production function would give a cubic-shape cost function?
user141240
# What kind of production function would give a cubic-shape cost function?

(Asked by: username_0 on 2021-11-01)

I would like a production function that gives a cost function with the following shape:

[![enter image description here](https://i.sstatic.net/Yv984.png)](https://i.sstatic.net/Yv984.png)

The figure was taken from "Microeconomic Theory: Basic Principles and Extensions, 12th edition", on Chapter 10, section 10.4.3

The cost function $c$ should satisfy the following requirements:

1. $c(q)$ is defined and continuous for all $q \geq 0$
2. $c(q)$ is twice continuously differentiable for all $q > 0$
3. $c(q) \to \infty$ as $q \to \infty$
4. For all $q > 0$, we have $c'(q) > 0$
5. There exists $q\_1 > 0$ such that $c''(q) < 0$ for all $0 < q < q\_1$ and $c''(q) > 0$ for all $q > q\_1$

I don't need $c$ to be literally a cubic polynomial (i.e. $ c(q) = a\_0 + a\_1 q + a\_2 q^2 + a\_3 q^3 $). I just need it to look like the shape in the figure and satisfy the above requirements.

**Tags:** microeconomics, production-function

> **Comments:**
> **username_1**: Do you require that $c(q) \rightarrow 0$ as $q \rightarrow 0$ (as in the picture)? If not, should be easy to do this using fixed costs. Also, do you want more than one factor input? It's quite trivial if you just have one input.  
> **username_0**: @username_1 Thank you for your comment. I don't need $c(0) = 0$ because as you said, it can be done with fixed costs. And yes, I want more then one input, otherwise it would be trivial.  

---

## 2 answers

### Answer by username_2 on 2021-11-01. Score: 2 | Accepted answer

As your cost function exhibits non constant first and second order derivative, a third degree polynomial is a good starting point.

There are infinitely many production functions behind a cubic cost function. Below there is just one example, using a trick from Gorman, which consists to interpret a cost function whose expression is the sum of three (or four) parts, as if it was obtained from a single firm using three machines (or three plants).  
Machine (or plant) number $j=1,...,3$, produces a constant share $s\_j$ of total output $y$ according to its own production function $h\_j$ such that $$s\_j y = h\_j(x\_j-k\_j),$$ where each $h\_j$ is homogeneous of degree $1/\alpha\_j$ in $(x\_j-k\_j)$, and $k\_j \geq 0$ denotes the minimum input requirement for production: $h\_j(x\_j-k\_j)=0$ for $x\_j \leq k\_j$. Output is denoted by $y$ and the input price vector by $w$ (it can be considered as constant here).
Then the optimal input demand vector takes the form (left as exercise):
$$x\_j^\*(w,y)=k\_j + b\_j(w)y^{\alpha\_j}$$
and the corresponding cost function is
$$c(w,y) = \sum\_{j=1}^3 w^T x\_j^\*(w,y) = a\_0(w) + a\_1(w)y^{\alpha\_1} + a\_2(w)y^{\alpha\_2} + a\_3(w)y^{\alpha\_3}$$
which is the expression of the cubic cost function for $\alpha\_1=1,\alpha\_2=2,$ and $\alpha\_3=3$.  
Further restrictions on the $a\_j$ function yield the above figures.

> **Comments:**
> **username_1**: I guess $k_j$ is the minimum amount of input $j$ required to produce anything? More generally, it might be good to define all the symbols in this answer (or at least provide a link).  
> **username_2**: @username_1: yes it is, I have added this point in the answer. As welle as some further explanations.  
> **username_1**: Thanks, that now makes more sense! Final question: can I just think of the $b_j(w)$ and $a_i(w)$ terms as positive constants?  
> **username_2**: Indeed, the functions $a_j$ and $b_j$ are constant in $y$ and are related by $a_j(w)=w^Tb_j(w)$ for $j=1,2,3$.  

---

### Answer by username_1 on 2021-11-01. Score: 1

As I noted in a comment, one approach is to allow for fixed costs (unlike in the graph). For example, suppose that production requires two factors $L$ and $K$ and that the function is symmetric Cobb-Douglas:
$$
q = L^\alpha K^\alpha \text{ where } \alpha \in (0, 1)
$$
Solving the firm's cost minimisation problem then [yields](http://www.econ.ucla.edu/sboard/teaching/econ11_09/econ11_09_slides7.pdf) the cost function
$$
C(q) = 2\sqrt{wr} q^{\frac{1}{2\alpha}} \equiv c q^{\frac{1}{2\alpha}}
$$
where $w$ and $r$ are the prices of $L$ and $K$ and $c>0$ is a constant that we've just defined. If we assume you also need to pay a fixed cost $F > 0$ to produce anything at all, then
$$
C(q)=
\begin{cases}
c q^{\frac{1}{2\alpha}} + F \text{ if } q > 0\\
0 \hspace{3em}\text{ if $q = 0$ }
\end{cases}
$$
If you choose $\alpha < 0.5$ (decreasing returns to scale), then you get a convex cost function with a discontinuity at zero. For example, if $\alpha = 0.25$, then costs start at $F$, fall discontinuously, and then increase quadratically in $q$. As a result, marginal costs are 'infinite' for the first unit, then drop discontinuously, and then rise as desired. Finally, average costs decrease and then rise. I realise that this isn't quite what you are after -- but it seems close.


---

