Overview
Today we cover costs before we put them together next class with revenues to solve the firm’s (first stage) profit maximization problem. While it seems we are adding quite a bit, and learning some new math problems with calculating costs, we will practice it more next class, when put together with revenues.
Readings
- Ch. 7 in Goolsbee, Levitt, and Syverson, 2019
Slides
Below, you can find the slides in two formats. Clicking the image will bring you to the html version of the slides in a new tab. Note while in going through the slides, you can type h to see a special list of viewing options, and type o for an outline view of all the slides.
The lower button will allow you to download a PDF version of the slides. I suggest printing the slides beforehand and using them to take additional notes in class (not everything is in the slides)!
Assignments
Problem Set 3 Due Mon Oct 18
Problem set 3 (on classes 2.1-2.3) is due by the end of the day Monday, October 18 by upload to Blackboard Assignments.
Appendix
The Relationship Between Returns to Scale and Costs
There is a direct relationship between a technology’s returns to scale1 and its cost structure: the rate at which its total costs increase2 and its marginal costs change3. This is easiest to see for a single input, such as our assumptions of the short run (where firms can change \(l\) but not \(\bar{k})\):
\[q=f(\bar{k},l)\]
Constant Returns to Scale:
Decreasing Returns to Scale
Increasing Returns to Scale
Cobb-Douglas Cost Functions
The total cost function for Cobb-Douglas production functions of the form \[q=l^{\alpha}k^{\beta}\] can be shown with some very tedious algebra to be:
\[C(w,r,q)=\left[\left(\frac{\alpha}{\beta}\right)^{\frac{\beta}{\alpha+\beta}} + \left(\frac{\alpha}{\beta}\right)^{\frac{-\alpha}{\alpha+\beta}}\right] w^{\frac{\alpha}{\alpha+\beta}} r^{\frac{\beta}{\alpha+\beta}} q^{\frac{1}{\alpha+\beta}}\]
If you take the first derivative of this (to get marginal cost), it is:
\[\frac{\partial C(w,r,q)}{\partial q}= MC(q) = \frac{1}{\alpha+\beta} \left(w^{\frac{\alpha}{\alpha+\beta}} r^{\frac{\beta}{\alpha+\beta}}\right) q^{\left(\frac{1}{\alpha+\beta}\right)-1}\]
How does marginal cost change with increased output? Take the second derivative:
\[\frac{\partial^2 C(w,r,q)}{\partial q^2}= \frac{1}{\alpha+\beta} \left(\frac{1}{\alpha+\beta} -1 \right) \left(w^{\frac{\alpha}{\alpha+\beta}} r^{\frac{\beta}{\alpha+\beta}}\right) q^{\left(\frac{1}{\alpha+\beta}\right)-2}\]
Three possible cases:
- If \(\frac{1}{\alpha+\beta} > 1\), this is positive \(\implies\) decreasing returns to scale
- Production function exponents \(\alpha+\beta < 1\)
- If \(\frac{1}{\alpha+\beta} < 1\), this is negative \(\implies\) increasing returns to scale
- Production function exponents \(\alpha+\beta > 1\)
- If \(\frac{1}{\alpha+\beta} = 1\), this is constant \(\implies\) constant returns to scale
- Production function exponents\(\alpha+\beta = 1\)
Example (Constant Returns)
Let \(q=l^{0.5}k^{0.5}\).
\[\begin{align*} C(w,r,q)&=\left[\left(\frac{0.5}{0.5}\right)^{\frac{0.5}{0.5+0.5}} + \left(\frac{0.5}{0.5}\right)^{\frac{-0.5}{0.5+0.5}}\right] w^{\frac{0.5}{0.5+0.5}} r^{\frac{0.5}{0.5+0.5}} q^{\frac{1}{0.5+0.5}}\\ C(w,r,q)&= \left[1^{0.5}+1^{-0.5} \right]w^{0.5}r^{0.5}q^{0.5}\\ C(w,r,q)&= w^{0.5}r^{0.5}q^{1}\\ \end{align*}\]
Consider input prices of \(w=\$9\) and \(r=\$25\):
\[\begin{align*}C(w=9,r=25,q)&=9^{0.5}25^{0.5}q \\ & =3*5*q\\ & =15q\\\end{align*}\]
That is, total costs (at those given input prices, and technology) is equal to 15 times the output level, \(q\):
Marginal costs would be
\[MC(q) = \frac{\partial C(q)}{\partial q} = 15\]
Average costs would be
\[MC(q) = \frac{C(q)}{q} = \frac{15q}{q} = 15\]
Example (Decreasing Returns)
Let \(q=l^{0.25}k^{0.25}\).
\[\begin{align*} C(w,r,q)&=\left[\left(\frac{0.25}{0.25}\right)^{\frac{0.25}{0.25+0.25}} + \left(\frac{0.25}{0.25}\right)^{\frac{-0.25}{0.25+0.25}}\right] w^{\frac{0.25}{0.25+0.25}} r^{\frac{0.25}{0.25+0.25}} q^{\frac{1}{0.25+0.25}}\\ C(w,r,q)&= \left[1^{0.5}+1^{-0.5} \right]w^{0.5}r^{0.5}q^{2}\\ C(w,r,q)&= w^{0.5}r^{0.5}q^{2}\\ \end{align*}\]
If \(w=9\), \(r=25\):
\[\begin{align*}C(w=9,r=25,q)&=9^{0.5}25^{0.5}q^2 \\ & =3*5*q^2\\ & =15q^2\\\end{align*}\]
Marginal costs would be
\[MC(q) = \frac{\partial C(q)}{\partial q} = 30q\]
Average costs would be
\[AC(q) = \frac{C(q)}{q} = \frac{15q^2}{q} = 15q\]
