# 2.4 — Costs of Production — Class Content

## Contents

- Section 1:
*Monday, March 15, 2021* - Section 2:
*Tuesday, March 24, 2021*Section 2, because of no class on Thursday Feb 18 and Tuesday Mar 9, is 2 classes behind Section 1.

## 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

## Assignments

## Appendix

### The Relationship Between Returns to Scale and Costs

There is a direct relationship between a technology’s returns to scaleIncreasing, decreasing, or constant

and its cost structure: the rate at which its total costs increaseAt a decreasing rate, at an increasing rate, or at a constant rate, respectively

and its marginal costs changeDecreasing, increasing, or constant, respectively

. 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

\[AC(q) = \frac{C(q)}{q} = \frac{15q^2}{q} = 15q\]