# CES Production Function with $\rho>1$

In using CES production functions of the form $f(x_1,x_2)=(x_1^\rho+x_2^\rho)^{1/\rho}$, we always assume that $\rho\leq1$. Why do we make that assumption? I understand that if $\rho>1$, the production function won't be concave anymore (and hence production set will not be convex), but what does that imply about profit and cost functions?

• $\rho$ above one would result in a corner solution where only one input is chosen with positive quantity. Since the point of multi-good production functions is usually to model circumstances where two inputs are actually used, this is an undesirable feature. – BKay Oct 25 '15 at 14:14
• Will there be a solution to profit max problem? – Sher Afghan Oct 25 '15 at 16:29
• @SherAfghan, linear function with $\rho = 1$ seems not to be in CES family, as its elasticity of substitution is not constant. – garej May 12 '18 at 5:06

The problem with $\rho>1$ is that it means the marginal product of factors is not decreasing ($\rho<1$) or constant ($\rho=1$) but increasing, which is an odd assumption. Such functions yield isoquants that are concave, and might lead to only one factor being used (as BKay said).

As in any generic CES, the marginal product of factor $x_i$ is

$$MP_i = \left(\frac{y}{x_i}\right)^{1-\rho}$$

The derivative of this MP with respect to $x_i$ is, after some rearranging,

$$(\rho-1) \left(\frac{y}{x_i}\right)^{1-\rho}\left(\frac{x_{-i}}{x_iy^{\rho}}\right)$$

For $\rho>1$, this expression is positive, which means that the productivity of a factor increases as more of that factor is used.

Regarding isoquants, you can find these by rewriting the production function as $x_2=g(y,x_1)$. In the generic CES, this is

$$x_2 = \left(y^{\rho} - x_1^{\rho}\right)^\frac{1}{\rho}$$

These are linear in the case of $\rho=1$, convex in the case of Cobb-Douglas (where the function above is $x_2=\frac{y}{x_1}$, a hyperbole), and concave in the case of $\rho>1$. For example, select $\rho=2$ and you have:

$$x_2^2 = y^2 - x_1^2$$

which is the formula of a circle centered at $(0,0)$, with radius $y$. Normally, for production theory only $x_i \geq 0$ is interesting, which gives you the concave isoquants for different levels of $y$. The figure below shows an example, were for a given factor prices ratio, there is a corner solution (point A):

$\hskip3cm$

(Code for reproducing figure here)

Here is my attempt at this question, it's incomplete and/or incorrect so please help make suggestions and I will edit this.

Cost Minimization

Since $f(x_1,x_2)$ is not quasi-concave, the corresponding isoquant curves are not going to be covex to the origin (i.e. their upper contour set will not be convex). In this case firm should employ corner solution and conditional factor demands will be given as; $$x_1(p,y)=q^2 \quad and \quad x_2(p,y)=0 \quad\quad if\quad w_1< w_2$$ $$x_1(p,y)=0 \quad and \quad x_2(p,y)=q^2 \quad\quad if\quad w_1>w_2$$ $$x_1(p,y)=0 , x_2(p,y)=q^2 \quad or \quad x_1(p,y)=q^2 , x_2(p,y)=0 \quad if\quad w_1=w_2$$ These conditional factor demands give the cost function; $$C(w,y)=min[w_1q^2,w_2q^2]$$ Profit Maximization

I am really confused here. Even though the production function is convex but it still exhibit non-increasing returns to scale. $f(tx_1,tx_2)<tf(x_1,x_2) \quad\forall \quad t>1$. That is the solution will still exist (right?). So how does non-concavity of production function effect profit maximizing solution?

• Your confusion is easy to clarify: Remember that convex preferences do not imply a concave utility function. They only imply that the upper contour sets are convex. Similarly, for the production function in question, consider $(x_1^{\rho}+x_2^{\rho})^{\theta/\rho}$. $\theta$ controls whether the function is concave or convex, $\rho$ controls whether the contour sets are convex. – HRSE Oct 26 '15 at 3:09
• I don't understand, concave function implies that the upper contour set are convex. $\rho<1$ means that the function is concave and that implies that it is quasi-concave, meaning the upper contour sets to the level set are convex. As far I understand $\theta$ in your example produce the monotonic transformation of original function which may or may not be concave. Anyways how does that effect profit maximizing solutions? – Sher Afghan Oct 27 '15 at 3:46
• Suppose $\rho\rightarrow \infty$. The above defined aggregate approaches the max function to the power of $\theta$. Thus, the upper contour sets are not convex. Now for every $\rho$ you can find a small enough $\theta$ such that the function has increasing or decreasing returns to scale. Returns to scale are thus unrelated to the convexity of the upper contour sets. – HRSE Oct 27 '15 at 4:21
• I see. So even if $\rho>1$, we can have the profit maximizing solution depending on value of $\theta$. Am I right in saying that we'll have solution (production function will exhibit decreasing returns to scale) if $\theta \leq 1$, on the other hand if $\theta >1$, the production function will exhibit increasing returns to scale and there will be no solution to profit max problem? – Sher Afghan Oct 27 '15 at 5:09
• Whether a solution to the profit maximization problem exists additionally depends on the market structure. A monopolist's profit maximization problem is usually still well defined, while for price-taking firms this will not be the case. – HRSE Oct 28 '15 at 1:19

In short, for $\rho \geq 1$ there going to be no solution for profit maximization in the short-run (at least one factor is fixed) for competitive case (price is fixes).

In order to get from production function to cost function, we need to introduce factor prices ($r$ and $w$ for textbooks examples) and solve optimization problem. Extensive exposition can be found here.

To build intuition, let's take $w=1$ and fix one factor. In order to deal with profit $\pi(q)$, we should introduce prices for produced goods as well $p>0$. So the problem might look as follows ($\rho=2$):

$$\pi(q) = p \cdot q - 1 \cdot (q^2 - 1)^{1/2}$$

It can be shown that for the profit function of this sort SOC is: $\pi'' >0$, which means there is no global maximum (though minimum exists).

To see the same effect in a simpler example (not derived from CES), consider this:

$$\pi(q) = p \cdot q - 2 \cdot q^{1/2}$$

SOC is $\pi'' = (1/2)q^{-3/2} > 0$.

Notice $q^{1/2}$ but not, say, $q^2$ as usual. Let us compare those two cases for $p=1.7$ on the plot to appreciate the difference.