Production Maximization with Lagrange Mutlipliers

In “Simple Optimization in 3D”, I blogged about a basic optimization problem in three dimensional space. In this post, I’ll look at a more complex problem that deals with an equation constraint. I’ll utilize the method of lagrange multipliers optimization strategy to solve the problem.

Code

library(plotly)
library(dplyr)

scene <- list(
  camera = list(eye = list(
    x = -2.2, y = 1.1, z = 1.2
  )),
  xaxis = list(title = "L"),
  yaxis = list(title = "K"),
  zaxis = list(title = "$")
)

The Optimization Problem

A company has determined that its production level is given by the Cobb-Douglas function $f(x,y)=2.5x^{0.45}y^{0.55}$ where $x$ represents the total number of labor hours in 1 year and $y$ represents the total capital input for the company. Suppose 1 unit of labor costs $40 and 1 unit of capital costs $50. Use the method of Lagrange multipliers to find the maximum value of $f(x,y)=2.5x^{0.45}y^{0.55}$ subject to a budgetary constraint of $500,000 per year.

– (Strang and Herman 2016, chap. 4.8)

\[ \begin{align} f(x, y) \rightarrow P(L, K) \\ P(L, K) &= 2.5L^{0.45}K^{0.55} \\ g(L, K) &= 40L + 50K - 500,000 \end{align} \tag{1}\]

We set the equation up in R so that we can inspect a plot and better understand the optimization problem.

P_l_k <- function(L, K) {
  2.5 * L ^ (0.45) * K ^ (0.55)
}
g_l_k <- function(L, K) {
  40 * L + 50 * K - 500000
}
n <- 100
L <- seq(0, 6000, length.out = n)
K <- seq(0, 6000, length.out = n)
P <- outer(L, K, P_l_k)
g <- outer(L, K, g_l_k)

Code

plot_ly(
  x = L,
  y = K,
  z = P,
  type = "surface",
  name = "P(L,K)"
) %>%
  colorbar(title = "P(L,K)") %>%
  add_trace(
    x = L,
    y = K,
    z = g,
    type = "surface",
    colorscale = "coolwarm",
    name = "g(L,K)",
    colorbar = list(title = "g(L,K)")
  ) %>% layout(scene = scene)

We see that the production function $P$ and the cost function $g$ are surfaces that intersect. We are looking for the highest possible point in $P$ that does not exceed the constraint $g$, which will be somewhere around their intersection. Note that generally, all values below the intersection are possible, though not profit-maximizing, points. The points higher than the intersection are more profit-maximizing, but are not possible with this budget constraint.

Maximizing using the Method of Lagrange Multipliers

We adapt the Lagrange multiplier problem-solving strategy from (Strang and Herman 2016, chap. 4.8) to our function input, and set up the following system of equations, which we will solve for $L_0$ and $K_0$:

\[ \begin{align*} \nabla P(L_0, K_0) &= \lambda \nabla g(L_0, K_0) \\ g(L_0, K_0) &= 0 \end{align*} \tag{2}\]

At this point, we will need to do some calculations to find each function in Equation 1’s gradient.

\[ \begin{align*} \nabla P(L_0, K_0) &= \left< \frac{1.125K^{0.55}}{L^{0.55}} , \frac{1.375L^{0.45}}{K^{0.45}}\right> \\ \nabla g(L_0, K_0) &= \left< 40, 50 \right> \\ \end{align*} \]

\[ \begin{align*} &\begin{cases} \left< \frac{1.125K^{0.55}}{L^{0.55}} , \frac{1.375L^{0.45}}{K^{0.45}}\right> &= \lambda \left< 40, 50 \right> \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} \frac{1.125K^{0.55}}{L^{0.55}} &= 40 \lambda\\ \frac{1.375L^{0.45}}{K^{0.45}} &= 50 \lambda \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} \frac{1.125K^{0.55}}{40L^{0.55}} &= \lambda\\ \frac{1.375L^{0.45}}{50K^{0.45}} &= \lambda \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} \frac{1.125K^{0.55}}{40L^{0.55}} &= \frac{1.375L^{0.45}}{50K^{0.45}} \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} 5.5L &= 5.625K \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} 5.5L- 5.625K &= 0 \\ 40L + 50K &= 500,000 \end{cases} \end{align*} \]

We now have a clear linear system of equations that we can solve via some substitution and algebraic manipulation:

\[ \begin{align*} &\begin{cases} L &= \frac{5.625K}{5.5} \\ 40 (\frac{5.625K}{5.5}) + 50K &= 500,000 \\ \end{cases} \\ &\begin{cases} L &= \frac{5.625K}{5.5} \\ K(40 (\frac{5.625}{5.5}) + 50) &= 500,000 \\ \end{cases} \\ &\begin{cases} L &= \frac{5.625K}{5.5} \\ K &= \frac{500,000}{(40 (\frac{5.625}{5.5}) + 50)} = 5,500 \end{cases} \\ &\begin{cases} L &= \frac{5.625 (5,500)}{5.5} = 5,625 \\ K &= 5,500 \end{cases} \\ &\begin{cases} L &= \boxed{5,625 \, \text{labor hours}} \\ K &= \boxed{\$ 5,500} \end{cases} \end{align*} \]

We’ll now plug those values for capital and labor into our production function and see how much output this maximizing parameter combination produces (we’ll round given we are solving for whole output):

P_l_k(5625, 5500) %>% round()

[1] 13890

When we return to the plot of the product function and budget constraint, we can see that this point clearly is the highest possible output under the constraints.

Code

plot_ly(
  x = L,
  y = K,
  z = P,
  type = "surface",
  name = "P(L,K)"
) %>%
  colorbar(title = "P(L,K)") %>%
  add_trace(
    x = L,
    y = K,
    z = g,
    type = "surface",
    colorscale = "coolwarm",
    name = "g(L,K)",
    colorbar = list(title = "g(L,K)")
  ) %>%
  add_trace(
    x = 5625,
    y = 5500,
    z = P_l_k(5625, 5500) %>% round(),
    type = "scatter3d",
    mode = "markers",
    marker = list(size = 5, color = "black")
  ) %>%
  layout(scene = scene, legend=list(x=.5, y=0))

However, in $R^3$, contour plots offer a much clearer way of visualizing our solution.

Code

plot_ly(
  x = L,
  y = K,
  z = P,
  type = "contour",
  name = "P(L,K)"
) %>%
  colorbar(title = "P(L,K)") %>%
  add_trace(
    x =  L,
    y = 10000 - 4 * K / 5,
    type = 'scatter',
    mode = 'lines',
    name = "g(L, K)",
    color = "red"
  ) %>%
  add_trace(
    x = 5625,
    y = 5500,
    type = "scatter",
    mode = "markers",
    marker = list(
      size = 10,
      color = "black",
      name = "P(L*,K*)"
    )
  ) %>%
  layout(xaxis = list(range = c(0, max(L))),
         yaxis = list(range = c(0, max(K))))

References

Strang, Gilbert, and Edwin Herman. 2016. Calculus Volume 3. OpenStax.

--- title: "Production Maximization with Lagrange Mutlipliers" bibliography: "../../blog.bib" author: "Peter Amerkhanian" date: "2024-3-17" categories: ['R', 'Calculus'] image: thumbnail.png format: html: toc: true toc-depth: 3 code-fold: false code-tools: true editor: markdown: wrap: 72 --- In ["Simple Optimization in 3D"](../minima-3d/index.qmd), I blogged about a basic optimization problem in three dimensional space. In this post, I'll look at a more complex problem that deals with an equation constraint. I'll utilize the [method of lagrange multipliers](https://en.wikipedia.org/wiki/Lagrange_multiplier) optimization strategy to solve the problem. ```{r} #| warning: false #| code-fold: true library(plotly) library(dplyr) scene <- list( camera = list(eye = list( x = -2.2, y = 1.1, z = 1.2 )), xaxis = list(title = "L"), yaxis = list(title = "K"), zaxis = list(title = "$") ) ``` ```{python} #| output: false #| echo: false import numpy as np import matplotlib.pyplot as plt # Define the functions def P_l_k(L, K): return 2.5 * L ** 0.45 * K ** 0.55 def g_l_k(L, K): return 40 * L + 50 * K - 500000 # Generate the data n = 100 L = np.linspace(0, 6000, n) K = np.linspace(0, 6000, n) L, K = np.meshgrid(L, K) P = P_l_k(L, K) g = g_l_k(L, K) # Create the contour plot fig, ax = plt.subplots(figsize=(3, 3)) # Plot the contour for P(L, K) contour = ax.contourf(L, K, P, cmap='Blues', alpha=.7) # Plot the line for g(L, K) L_line = np.linspace(0, 6000, n) K_line = (10000 - 4 * L_line / 5) # Add the point (5625, 5500) #ax.plot(L_line, K_line, color='black', linewidth="3", zorder=1) #ax.scatter(5625, 5500, color='black', s=50, zorder=4) # Set axis ranges and labels ax.set_xlim(0, 6000) ax.set_ylim(0, 6000) ax.set_axis_off() fig.tight_layout() fig.savefig("thumbnail.png", dpi=300) # Show the plot plt.show() ``` ### The Optimization Problem > A company has determined that its production level is given by the > Cobb-Douglas function $f(x,y)=2.5x^{0.45}y^{0.55}$ where $x$ > represents the total number of labor hours in 1 year and $y$ > represents the total capital input for the company. Suppose 1 unit of > labor costs \$40 and 1 unit of capital costs \$50. Use the method of > Lagrange multipliers to find the maximum value of > $f(x,y)=2.5x^{0.45}y^{0.55}$ subject to a budgetary constraint of > \$500,000 per year. > > -- [@strang_calculus_2016-1, chapter 4.8] $$ \begin{align} f(x, y) \rightarrow P(L, K) \\ P(L, K) &= 2.5L^{0.45}K^{0.55} \\ g(L, K) &= 40L + 50K - 500,000 \end{align} $$ {#eq-all} We set the equation up in `R` so that we can inspect a plot and better understand the optimization problem. ```{r} P_l_k <- function(L, K) { 2.5 * L ^ (0.45) * K ^ (0.55) } g_l_k <- function(L, K) { 40 * L + 50 * K - 500000 } n <- 100 L <- seq(0, 6000, length.out = n) K <- seq(0, 6000, length.out = n) P <- outer(L, K, P_l_k) g <- outer(L, K, g_l_k) ``` ```{r} #| code-fold: true plot_ly( x = L, y = K, z = P, type = "surface", name = "P(L,K)" ) %>% colorbar(title = "P(L,K)") %>% add_trace( x = L, y = K, z = g, type = "surface", colorscale = "coolwarm", name = "g(L,K)", colorbar = list(title = "g(L,K)") ) %>% layout(scene = scene) ``` We see that the production function $P$ and the cost function $g$ are surfaces that intersect. We are looking for the highest possible point in $P$ that does not exceed the constraint $g$, which will be somewhere around their intersection. Note that generally, all values below the intersection are possible, though not profit-maximizing, points. The points higher than the intersection are more profit-maximizing, but are not possible with this budget constraint. ### Maximizing using the Method of Lagrange Multipliers We adapt the Lagrange multiplier problem-solving strategy from [@strang_calculus_2016-1, chapter 4.8] to our function input, and set up the following system of equations, which we will solve for $L_0$ and $K_0$: $$ \begin{align*} \nabla P(L_0, K_0) &= \lambda \nabla g(L_0, K_0) \\ g(L_0, K_0) &= 0 \end{align*} $$ {#eq-lagrange} At this point, we will need to do some calculations to find each function in @eq-all's gradient. $$ \begin{align*} \nabla P(L_0, K_0) &= \left< \frac{1.125K^{0.55}}{L^{0.55}} , \frac{1.375L^{0.45}}{K^{0.45}}\right> \\ \nabla g(L_0, K_0) &= \left< 40, 50 \right> \\ \end{align*} $$ $$ \begin{align*} &\begin{cases} \left< \frac{1.125K^{0.55}}{L^{0.55}} , \frac{1.375L^{0.45}}{K^{0.45}}\right> &= \lambda \left< 40, 50 \right> \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} \frac{1.125K^{0.55}}{L^{0.55}} &= 40 \lambda\\ \frac{1.375L^{0.45}}{K^{0.45}} &= 50 \lambda \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} \frac{1.125K^{0.55}}{40L^{0.55}} &= \lambda\\ \frac{1.375L^{0.45}}{50K^{0.45}} &= \lambda \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} \frac{1.125K^{0.55}}{40L^{0.55}} &= \frac{1.375L^{0.45}}{50K^{0.45}} \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} 5.5L &= 5.625K \\ 40L + 50K - 500,000 &= 0 \end{cases} \\ &\begin{cases} 5.5L- 5.625K &= 0 \\ 40L + 50K &= 500,000 \end{cases} \end{align*} $$ We now have a clear linear system of equations that we can solve via some substitution and algebraic manipulation: $$ \begin{align*} &\begin{cases} L &= \frac{5.625K}{5.5} \\ 40 (\frac{5.625K}{5.5}) + 50K &= 500,000 \\ \end{cases} \\ &\begin{cases} L &= \frac{5.625K}{5.5} \\ K(40 (\frac{5.625}{5.5}) + 50) &= 500,000 \\ \end{cases} \\ &\begin{cases} L &= \frac{5.625K}{5.5} \\ K &= \frac{500,000}{(40 (\frac{5.625}{5.5}) + 50)} = 5,500 \end{cases} \\ &\begin{cases} L &= \frac{5.625 (5,500)}{5.5} = 5,625 \\ K &= 5,500 \end{cases} \\ &\begin{cases} L &= \boxed{5,625 \, \text{labor hours}} \\ K &= \boxed{\$ 5,500} \end{cases} \end{align*} $$ We'll now plug those values for capital and labor into our production function and see how much output this maximizing parameter combination produces (we'll round given we are solving for whole output): ```{r} P_l_k(5625, 5500) %>% round() ``` When we return to the plot of the product function and budget constraint, we can see that this point clearly is the highest possible output under the constraints. ```{r} #| code-fold: true plot_ly( x = L, y = K, z = P, type = "surface", name = "P(L,K)" ) %>% colorbar(title = "P(L,K)") %>% add_trace( x = L, y = K, z = g, type = "surface", colorscale = "coolwarm", name = "g(L,K)", colorbar = list(title = "g(L,K)") ) %>% add_trace( x = 5625, y = 5500, z = P_l_k(5625, 5500) %>% round(), type = "scatter3d", mode = "markers", marker = list(size = 5, color = "black") ) %>% layout(scene = scene, legend=list(x=.5, y=0)) ``` However, in $R^3$, contour plots offer a much clearer way of visualizing our solution. ```{r} #| warning: false #| code-fold: true plot_ly( x = L, y = K, z = P, type = "contour", name = "P(L,K)" ) %>% colorbar(title = "P(L,K)") %>% add_trace( x = L, y = 10000 - 4 * K / 5, type = 'scatter', mode = 'lines', name = "g(L, K)", color = "red" ) %>% add_trace( x = 5625, y = 5500, type = "scatter", mode = "markers", marker = list( size = 10, color = "black", name = "P(L*,K*)" ) ) %>% layout(xaxis = list(range = c(0, max(L))), yaxis = list(range = c(0, max(K)))) ```