10  Production Part 1

Recommended reading: Allen, Doherty, Weigelt, and Mansfield Chapter 5: Production Theory

10.1 Classwork 10

  1. Consider the production function below. Plot labor on the x-axis and output on the y-axis. As you keep increasing labor, what is one reason that output might begin to fall?
library(tidyverse)

production <- tibble(
    labor = 0:15,
    capital = rep(5, 16),
    output = c(0, 49, 132, 243, 376, 525, 684, 847, 1008, 1161, 1300, 1419, 1512, 1573, 1596, 1575)
    )

  1. Take the production function data and use lm() to estimate the model \(\text{output} = \beta_0 + \beta_1 \text{labor} + \beta_2 \text{labor}^2 + \text{labor}^3 + u\). Show that the results are that production is given by \(Q = 30 L + 20 L^2 - L^3\).

  2. Edit the variable definition of production to include a column average_product that is equal to output divided by labor. Then plot average product on the y-axis against labor on the x-axis. For what amount of labor does average product reach its maximum?

  3. If the production function is given by \(Q = 30 L + 20 L^2 - L^3\), then the average product of labor is given by \(\frac{Q}{L} = \underline{\hspace{2cm}}\). Find the average product’s maximum by finding where the derivative of the average product is equal to 0 (and verify that it’s a maximum instead of a minimum by looking at the second derivative).

  4. Edit the variable definition of production to include a column marginal_product that is equal to how the last worker affected the firm’s output. Hint: you might find that lag() is useful here: it gives the previous value for a variable. To see what it does, try: mutate(production, lag_output = lag(output)) When does the marginal product reach its maximum?

  5. If the production function is given by \(Q = 30 L + 20 L^2 - L^3\), then the marginal product of labor is given by \(\frac{dQ}{dL} = \underline{\hspace{2cm}}\). If fractional amounts of labor are attainable (part time or overtime), find the amount of labor that maximizes the marginal product (find where the derivative of the marginal product is equal to 0, and look at the second derivative to verify you’ve reached a maximum).

  6. One important fact about production functions is that the average product will always be maximized when the marginal product is equal to the average product. Show that this rule holds true for this dataset using a plot of average product and marginal product (in different colors).

  7. The intuition behind this “When AP is maximized, MP = AP” rule is this: say I’m grading exams and I’m keeping a running average: 82, then 76, then 83. If my running average is 83 again, what do you know about the test I must have just graded? You know it had to have been an 83. In the production function example, say average product is 80 for 1 worker, then 95 for 2 workers, then 100 for 3 workers, then 100 again for 4 workers, and then it starts falling. What do you know about the contribution of the 4th worker? It had to have been 100: that’s the only way to get an average product of 100 twice in a row. Therefore when average product is maximized (100 in this example), we’ll get that the marginal product is 100, which is equal to that maximum average product. Finish the proof that this rule holds in the general case.

We want to show that AP = MP when AP is maximized. That is, for the value of L where AP is maximized, we have \(\frac{dQ}{dL} = \frac{Q}{L}\).

Start with the fact that the first derivative of AP is equal to zero when AP is maximized. That is,

\[\frac{d(Q/L)}{dL} = 0\] Remember that Q is a function of L:

\[\frac{d(Q(L) L^{-1})}{dL} = 0\]

Take the derivative, applying the product rule:

\[\underline{\hspace{2cm}} = 0\] Simplify to see we are done: we’ve showed that where the average product is maximized, the marginal product is equal to the average product.

\[\frac{dQ}{dL} = \frac{Q}{L}\]