/*! 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 34 Show that the Cobb-Douglas funct... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that the Cobb-Douglas function $$ Q=b K^{\alpha} L^{1-\alpha} \text { where } \quad 0 < \alpha < 1 $$ satisfies the equation $$ K \frac{\partial Q}{\partial K}+L \frac{\partial Q}{\partial L}=Q $$

Short Answer

Expert verified
The Cobb-Douglas function satisfies the equation by confirming the identities for partial derivatives and simplification.

Step by step solution

01

Compute Partial Derivative with respect to K

The Cobb-Douglas function is given by \( Q = b K^{\alpha} L^{1-\alpha} \). First, find the partial derivative of \( Q \) with respect to \( K \). Differentiating gives:\[ \frac{\partial Q}{\partial K} = \frac{\partial}{\partial K} (b K^{\alpha} L^{1-\alpha}) = b \alpha K^{\alpha-1} L^{1-\alpha}. \]
02

Compute Partial Derivative with respect to L

Next, compute the partial derivative of \( Q \) with respect to \( L \). Differentiating gives:\[ \frac{\partial Q}{\partial L} = \frac{\partial}{\partial L} (b K^{\alpha} L^{1-\alpha}) = b (1-\alpha) K^{\alpha} L^{-\alpha}. \]
03

Substitute Partial Derivatives into Given Equation

Substitute \( \frac{\partial Q}{\partial K} \) and \( \frac{\partial Q}{\partial L} \) into the equation \( K \frac{\partial Q}{\partial K} + L \frac{\partial Q}{\partial L} \):\[K \frac{\partial Q}{\partial K} + L \frac{\partial Q}{\partial L} = K (b \alpha K^{\alpha-1} L^{1-\alpha}) + L (b (1-\alpha) K^{\alpha} L^{-\alpha}).\]
04

Simplify the Expression

Simplify both terms in the expression:\[ K \cdot b \alpha K^{\alpha-1} L^{1-\alpha} = b \alpha K^{\alpha} L^{1-\alpha},\]and\[L \cdot b (1-\alpha) K^{\alpha} L^{-\alpha} = b (1-\alpha) K^{\alpha} L^{1-\alpha}.\] Combine these:\[ K \frac{\partial Q}{\partial K} + L \frac{\partial Q}{\partial L} = b \alpha K^{\alpha} L^{1-\alpha} + b (1-\alpha) K^{\alpha} L^{1-\alpha}.\]
05

Conclude with Algebraic Identity

Factor the expression:\[K \frac{\partial Q}{\partial K} + L \frac{\partial Q}{\partial L} = b K^{\alpha} L^{1-\alpha} (\alpha + (1-\alpha)) = b K^{\alpha} L^{1-\alpha} \cdot 1 = Q.\]Thus, it matches the original function \( Q \), proving the given equation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Partial Derivatives
In differential calculus, partial derivatives are a way of understanding how a multivariable function changes as each variable changes independently. For example, in the Cobb-Douglas production function, each variable represents a factor of production, like capital (\(K\)) and labor (\(L\)). By taking the partial derivative of the function with respect to one of these variables, we can see how the output \(Q\) changes when only that particular input changes.

For the Cobb-Douglas function \( Q = b K^{\alpha} L^{1-\alpha} \), when taking the partial derivative with respect to \(K\), we treat \(L\) as a constant. This yields \( \frac{\partial Q}{\partial K} = b \alpha K^{\alpha-1} L^{1-\alpha} \).

Similarly, the partial derivative with respect to \(L\) is found by treating \(K\) as a constant, resulting in \( \frac{\partial Q}{\partial L} = b (1-\alpha) K^{\alpha} L^{-\alpha} \). These derivatives provide insight into the responsiveness of output to changes in each input.
Differential Calculus
Differential calculus is the mathematical tool used for understanding change. It involves derivatives, which measure how a function's output varies as its input changes. This is crucial in economics to determine how varying levels of input such as capital and labor impact production levels.

In the context of the Cobb-Douglas function, differential calculus helps us derive the expressions for partial derivatives, as demonstrated previously. These derivatives are instrumental in analyzing shifts in production output based on small changes in inputs.

By employing the approach of differential calculus on the Cobb-Douglas function, we can verify critical production properties like the efficiency of input combinations and returns to scale. This helps businesses optimize their operational processes, ensuring maximum efficiency.
Output Elasticity
Output elasticity is a measure of how sensitive the output is to a change in one of the inputs. Each partial derivative of the Cobb-Douglas function with respect to a factor of production indicates the elasticity of output concerning that input.

For example, the partial derivative of the Cobb-Douglas function with respect to \(K\), given by \( \frac{\partial Q}{\partial K} = b \alpha K^{\alpha-1} L^{1-\alpha} \), reveals how a 1% change in capital will affect the output in percentage terms. The coefficient \(\alpha\) directly represents this elasticity. Similarly, \(1-\alpha\) shows the output elasticity concerning labor \(L\).

Understanding these elasticities allows firms to anticipate changes in output when allocating resources differently, guiding strategic decisions in production planning and resource management.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sketch a contour diagram for the function with at least four labeled contours. Describe in words the contours and how they are spaced. $$f(x, y)=x^{2}+y^{2}$$

A firm manufactures a commodity at two different factories. The total cost of manufacturing depends on the quantities, \(q_{1}\) and \(q_{2},\) supplied by each factory, and is expressed by the joint cost function, $$C=f\left(q_{1}, q_{2}\right)=2 q_{1}^{2}+q_{1} q_{2}+q_{2}^{2}+500$$ The company's objective is to produce 200 units, while minimizing production costs. How many units should be supplied by each factory?

A steel manufacturer can produce \(P(K, L)\) tons of steel using \(K\) units of capital and \(L\) units of labor, with production costs \(C(K, L)\) dollars. With a budget of \(\$ 600,000,\) the maximum production is 2,500,000 tons, using \(\$ 400,000\) of capital and \(\$ 200,000\) of labor. The Lagrange multiplier is \(\lambda=3.17\) (a) What is the objective function? (b) What is the constraint? (c) What are the units for \(\lambda ?\) (d) What is the practical meaning of the statement \(\lambda=\) \(3.17 ?\)

Problems are about the money supply, \(M,\) which is the total value of all the cash and checking account balances in an economy. It is determined by the value of all the cash, \(B\), the ratio, \(c,\) of cash to checking deposits, and the fraction, \(r,\) of checking account deposits that banks hold as cash: $$ M=\frac{c+1}{c+r} B $$ (a) Find the partial derivative. (b) Give its sign. (c) Explain the significance of the sign in practical terms. $$\partial M / \partial r$$

For a function \(f(x, y),\) we are given \(f(100,20)=2750\) and \(f_{x}(100,20)=4,\) and \(f_{y}(100,20)=7 .\) Estimate \(f(105,21)\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.