/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 The output of an economic system... [FREE SOLUTION] | 91Ó°ÊÓ

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The output of an economic system \(Q,\) subject to two inputs, such as labor \(L\) and capital \(K\) is often modeled by the Cobb-Douglas production function \(Q=c L^{a} K^{b} .\) When \(a+b=1,\) the case is called constant returns to scale. Suppose \(Q=1280, a=\frac{1}{3}, b=\frac{2}{3},\) and \(c=40\) a. Find the rate of change of capital with respect to labor, \(d K / d L\). b. Evaluate the derivative in part (a) with \(L=8\) and \(K=64\)

Short Answer

Expert verified
Answer: The rate of change of capital with respect to labor (\(\frac{dK}{dL}\)) when \(L = 8\) is -16.

Step by step solution

01

Substitute the given values into the production function

We are given the Cobb-Douglas production function \(Q = cL^aK^b\), with \(Q = 1280\), \(a = \frac{1}{3}\), \(b = \frac{2}{3}\), and \(c = 40\). We can plug in these values into the function: \(1280 = 40L^{\frac{1}{3}}K^{\frac{2}{3}}\)
02

Solve the production function for K

We want to solve the production function for K. To do this, let's divide both sides of the equation by \(40L^{\frac{1}{3}}\): \(K^{\frac{2}{3}} = \frac{1280}{40L^{\frac{1}{3}}}\) Now, raise both sides to the power of \(\frac{3}{2}\) to isolate K: \(K = \left(\frac{1280}{40L^{\frac{1}{3}}}\right)^{\frac{3}{2}}\)
03

Differentiate K with respect to L

Now, we need to find \(\frac{dK}{dL}\). Differentiating the equation for K with respect to L, we get: \(\frac{dK}{dL} = \frac{3}{2} \left(\frac{1280}{40}\right)^{\frac{3}{2}} L^{-\frac{1}{2}}(\frac{-1}{3})\) Simplify the expression: \(\frac{dK}{dL} = -\frac{1}{2} \left(\frac{1280}{40}\right)^{\frac{3}{2}} L^{-\frac{1}{2}}\)
04

Evaluate the derivative at the given values

We are given that \(L = 8\) and \(K = 64\). Plug in these values into the expression we found for \(\frac{dK}{dL}\): \(\frac{dK}{dL}\Big|_{L=8} = -\frac{1}{2} \left(\frac{1280}{40}\right)^{\frac{3}{2}} (8)^{-\frac{1}{2}}\) Calculate the result: \(\frac{dK}{dL}\Big|_{L=8} = -\frac{1}{2} (32)^{\frac{3}{2}} (8)^{-\frac{1}{2}} = -16\) After following these steps, we have our answers: a. The rate of change of capital with respect to labor (\(\frac{dK}{dL}\)) is given by \(\frac{dK}{dL} = -\frac{1}{2} \left(\frac{1280}{40}\right)^{\frac{3}{2}} L^{-\frac{1}{2}}\). b. Evaluating the derivative in part (a) at \(L = 8\), we find that \(\frac{dK}{dL}\Big|_{L=8} = -16\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Production Function
In economics, a production function reflects the relationship between input quantities and the output produced. The Cobb-Douglas production function is one of the most widely known mathematical models. It's characterized by its formula:
  • \(Q = c L^a K^b\): Where \(Q\) represents the total output, \(L\) the input of labor, \(K\) the input of capital, and \(c\), \(a\), and \(b\) are constants with \(a + b = 1\) for constant returns to scale.
This function typically features diminishing marginal returns for inputs, meaning each additional unit of input contributes less to output than the previous unit did.
In our exercise, we have specific values: \(Q = 1280\), \(a = \frac{1}{3}\), \(b = \frac{2}{3}\), and \(c = 40\). Understanding this function helps predict how variations in labor and capital affect output.
Differentiation
Differentiation is a mathematical process used to find the rate at which one variable changes relative to another. In the context of production functions, we use differentiation to determine how changes in labor impact capital or output.
For our Cobb-Douglas function, differentiating helps find the rate of change of capital \(K\) with respect to labor \(L\), represented by \(\frac{dK}{dL}\). This involves applying techniques of calculus to the function and requires an understanding of how to manipulate equations to solve for different variables.
In this exercise, we applied differentiation to the expression for \(K\) derived from the production function to find \(\frac{dK}{dL} = -\frac{1}{2} \left(\frac{1280}{40}\right)^{\frac{3}{2}} L^{-\frac{1}{2}}\). This tells us about the sensitivity of capital in response to changes in labor input.
Constant Returns to Scale
Constant returns to scale is a concept where increasing all inputs by a certain proportion leads to an equal proportionate increase in output. This property is visible in the Cobb-Douglas function when \(a + b = 1\), creating a linear scaling of inputs to outputs.
  • For example, if labor and capital inputs are each doubled, the total output \(Q\) also doubles.
In practice, constant returns to scale are pivotal for analyzing an organization’s ability to expand production without becoming inefficient or overly costly.
In our exercise, the sum of exponents \(a = \frac{1}{3}\) and \(b = \frac{2}{3}\) adds up to 1, confirming the presence of constant returns to scale.
Rate of Change
The rate of change in a mathematical sense often describes how one quantity changes in relation to another. For the Cobb-Douglas production function, the rate of change we are interested in is \(\frac{dK}{dL}\), the derivative of capital with respect to labor.
  • This measurement indicates how much capital needs to be adjusted for a change in labor while maintaining the same level of output.
  • A negative rate, for instance, shows that an increase in labor results in a reduction in capital to keep output constant.
In the specific problem, once evaluated with \(L = 8\) and \(K = 64\), the expression \(\frac{dK}{dL}\) results in -16. This specific value shows that for every unit increase in labor, capital must decrease by 16 units to sustain the production level of 1280.

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Most popular questions from this chapter

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