<p><strong>A note on the parameter constraints:</strong>
$$TC(q)=a+bq+cq^{2}+dq^{3} = a + q(b+cq+dq^{2})$$
It appears that what we want is that for the second degree polynomial to not take negative values. This would require for it to not have real and positive roots. Since $c&lt;0$, then if there are real roots one of them will be positive. This would imply that for an interval of positive values of $q$ the 2nd degree polynomial will take on negative values, and marginal cost will turn negative. So we need the discriminant to be negative, and so $c^2 - 4bd &lt; 0 \implies c^2 &lt; 4bd$. I don't see whey they write "$3$" instead of "$4$".  </p>

<p>-__________________________________</p>

<p>As for obtaining such a cost function rigorously from a production function:<br>
The reference is <strong><a href="https://books.google.gr/books/about/The_Structure_of_Economics.html?id=DuW6AAAAIAAJ&amp;redir_esc=y" rel="nofollow">Silberberg. E (1990), "The Structure of Economics" (2nd ed)</a></strong>, ch. 9.  </p>

<p><strong>A)</strong> <strong>When the production function is homogeneous of degree</strong> $r$, then the cost function has the form</p>

<p>$$C(q,\mathbb w) = q^{1/r} \cdot h(\mathbb w)$$</p>

<p>where $h(\mathbb w)$ is a function of prices (of input factors), and it is homogeneous of degree one (or "linearly homogeneous").</p>

<p><strong>B) When the production function is homothetic</strong>, which can be represented as a monotonic function of a homogeneous of degree one function, then the cost function remains multiplicatively separable in output and prices for some function $J(q)$:</p>

<p>$$C(q,\mathbb w) = J(q)\cdot h(\mathbb w)$$</p>

<p>But again, since homotheticity is a monotonic transformation of homogeneity of degree one, $J(q)$ should not be expected to take any polynomial form like the one you seek.  </p>

<p>So none of the usual functional specifications for production functions will give a cost function like the one you want to arrive at.  And indeed in papers with such cost functions I have never seen a derivation of the underlying production function.  </p>

<p>If any positive result comes up, I will return.</p>

CC BY-SA 3.0
2015-06-08T23:39:06.673Z
2015-06-08T23:39:06.673Z
Alecos Papadopoulos
<p>Given a Cobb-Douglas production function your cost function will be cubic exactly if $\alpha + \beta = \frac{1}{3}$, and in this case you will have $a = b = c = 0$. There is no going around this. If you want to have the kind of cost function you describe you will have to slightly alter your Cobb-Douglas function, e.g.
$$
f(k,l) = (k-1)^{\alpha} \cdot (l-2)^{\frac{1}{3} - \alpha}.
$$</p>

CC BY-SA 3.0
2015-06-06T23:35:58.093Z
2015-06-06T23:35:58.093Z
Giskard
<p>The cubic total cost function usually take the form</p>

<p>$TC(q)=a+bq+cq^{2}+dq^{3} \qquad a,b,d&gt;0, c&lt;0$ and $c^{2}&lt;4bd$</p>

<p>I know that from a constraint maximization problem </p>

<p>$min\quad wL+vK$ </p>

<p>subject to </p>

<p>$q_{0}=f(k,l)$</p>

<p>i can express it with the lagrangean function </p>

<p>$\mathcal{L}=wl+vk+\lambda(q_{0}-f(k,l))$</p>

<p>with some algebra for the case of a Cobb-Douglas production function $q_{0}=k^{\alpha}l^{\beta}$ i can reach </p>

<p>$TC=q^{\frac{1}{\alpha+\beta}}w^{\frac{\beta}{\alpha+\beta}}v^{\frac{\alpha}{\alpha+\beta}}\left(\frac{\alpha+\beta}{\alpha^{\frac{\alpha}{\alpha+\beta}}\beta^{\frac{\beta}{\alpha+\beta}}}\right)$</p>

<p>I could give values to $\alpha$ and $\beta$ to obtain a cubic function, but not the one I described in the preamble. It could also fix some other factor such as the size of the factory, but would not work.</p>

<p>Any idea? </p>

<p>Thanks in advance</p>

CC BY-SA 3.0
2015-06-06T23:37:48.703Z
You write $TC(q)$ usually takes this very specific form. Can you please post a reference/source? I have never seen these parameter restrictions before.
Giskard
CC BY-SA 3.0
2015-06-06T23:48:49.743Z
The texts do not usually specify this. But, in practice, these restrictions $a,b,c>0, \quad c<0$ and $c^{2}<3bd$ make sense to not have solutions in the field of complex numbers. a text that uses this (page 2): http://ijecm.co.uk/wp-content/uploads/2014/02/2214.pdf
Héctor Garrido
<microeconomics><production-function><cost>
How to derive cubic cost function from a problem of constrained optimization?
Héctor Garrido
# How to derive cubic cost function from a problem of constrained optimization?

(Asked by: username_0 on 2015-06-06)

The cubic total cost function usually take the form

$TC(q)=a+bq+cq^{2}+dq^{3} \qquad a,b,d>0, c<0$ and $c^{2}<4bd$

I know that from a constraint maximization problem

$min\quad wL+vK$

subject to

$q\_{0}=f(k,l)$

i can express it with the lagrangean function

$\mathcal{L}=wl+vk+\lambda(q\_{0}-f(k,l))$

with some algebra for the case of a Cobb-Douglas production function $q\_{0}=k^{\alpha}l^{\beta}$ i can reach

$TC=q^{\frac{1}{\alpha+\beta}}w^{\frac{\beta}{\alpha+\beta}}v^{\frac{\alpha}{\alpha+\beta}}\left(\frac{\alpha+\beta}{\alpha^{\frac{\alpha}{\alpha+\beta}}\beta^{\frac{\beta}{\alpha+\beta}}}\right)$

I could give values to $\alpha$ and $\beta$ to obtain a cubic function, but not the one I described in the preamble. It could also fix some other factor such as the size of the factory, but would not work.

Any idea?

Thanks in advance

**Tags:** microeconomics, production-function, cost
> **Comments:**
> **username_1**: You write $TC(q)$ usually takes this very specific form. Can you please post a reference/source? I have never seen these parameter restrictions before.  
> **username_0**: The texts do not usually specify this. But, in practice, these restrictions $a,b,c>0, \quad c<0$ and $c^{2}<3bd$ make sense to not have solutions in the field of complex numbers. a text that uses this (page 2): http://ijecm.co.uk/wp-content/uploads/2014/02/2214.pdf  

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## 2 answers

### Answer by username_2 on 2015-06-08. Score: 2 | Accepted answer

**A note on the parameter constraints:**
$$TC(q)=a+bq+cq^{2}+dq^{3} = a + q(b+cq+dq^{2})$$
It appears that what we want is that for the second degree polynomial to not take negative values. This would require for it to not have real and positive roots. Since $c<0$, then if there are real roots one of them will be positive. This would imply that for an interval of positive values of $q$ the 2nd degree polynomial will take on negative values, and marginal cost will turn negative. So we need the discriminant to be negative, and so $c^2 - 4bd < 0 \implies c^2 < 4bd$. I don't see whey they write "$3$" instead of "$4$".

-\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

As for obtaining such a cost function rigorously from a production function:  
The reference is **[Silberberg. E (1990), "The Structure of Economics" (2nd ed)](https://books.google.gr/books/about/The_Structure_of_Economics.html?id=DuW6AAAAIAAJ&redir_esc=y)**, ch. 9.

**A)** **When the production function is homogeneous of degree** $r$, then the cost function has the form

$$C(q,\mathbb w) = q^{1/r} \cdot h(\mathbb w)$$

where $h(\mathbb w)$ is a function of prices (of input factors), and it is homogeneous of degree one (or "linearly homogeneous").

**B) When the production function is homothetic**, which can be represented as a monotonic function of a homogeneous of degree one function, then the cost function remains multiplicatively separable in output and prices for some function $J(q)$:

$$C(q,\mathbb w) = J(q)\cdot h(\mathbb w)$$

But again, since homotheticity is a monotonic transformation of homogeneity of degree one, $J(q)$ should not be expected to take any polynomial form like the one you seek.

So none of the usual functional specifications for production functions will give a cost function like the one you want to arrive at. And indeed in papers with such cost functions I have never seen a derivation of the underlying production function.

If any positive result comes up, I will return.


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### Answer by username_1 on 2015-06-06. Score: 2

Given a Cobb-Douglas production function your cost function will be cubic exactly if $\alpha + \beta = \frac{1}{3}$, and in this case you will have $a = b = c = 0$. There is no going around this. If you want to have the kind of cost function you describe you will have to slightly alter your Cobb-Douglas function, e.g.
$$
f(k,l) = (k-1)^{\alpha} \cdot (l-2)^{\frac{1}{3} - \alpha}.
$$


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