Problem 47 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

The output at a certain factory is \(Q(L)=600 L^{2 / 3}\) units, where \(L\) is the size of the labor force. The manufacturer wishes to increase output by \(1 \%\). Use calculus to estimate the percentage increase in labor that will be required.

Step-by-Step Solution

Answer

A 1.5% increase in labor is required to increase the output by 1%.

Step 1: Define Variables and Function

Define the output function as given: \[ Q(L) = 600L^{2/3} \] Here, \( Q(L) \) represents the output units, and \(L\) is the size of the labor force.

Step 2: Differentiate the Function

Find the derivative of the output function with respect to labor \( L \). \[ \frac{dQ}{dL} = 600 \cdot \frac{2}{3} L^{-1/3} = 400 L^{-1/3} \] This represents the rate of change of output with respect to labor force.

Step 3: Calculate the Desired Change in Output

Given that the manufacturer wishes to increase the output by 1%, calculate the desired change in output: \[ dQ = 0.01 \times Q(L) \]

Step 4: Use Differentiate Approximation

Using the differentiable approximation, approximate the change in labor required: \[ dQ \approx \frac{dQ}{dL} dL \] Substitute the values found: \[ 0.01 \times Q(L) \approx 400 L^{-1/3} dL \]

Step 5: Solve for Change in Labor \(dL\)

Rearrange the equation to solve for \( dL \): \[ dL \approx \frac{0.01 \times Q(L)}{400 L^{-1/3}} = \frac{0.01 \times 600 L^{2/3}}{400 L^{-1/3}} = \frac{0.01 \times 600}{400} L = \frac{3}{200} L \]

Step 6: Convert to Percentage

Convert the calculated change in labor to a percentage increase: \[ \frac{dL}{L} \times 100 = \frac{3}{200} \times 100 = 1.5\% \]

Key Concepts

Derivatives

In calculus, derivatives represent the rate at which a function is changing at any given point. For our exercise, we're interested in how the output changes with respect to the size of the labor force. The given output function is \(Q(L) = 600L^{2/3}\). By differentiating this function with respect to \(L\), we obtain the derivative \(\frac{dQ}{dL} = 400L^{-1/3}\). This tells us how a small change in labor \(L\) will affect the output \(Q\). Derivatives are crucial in economics for analyzing how changes in one variable, such as labor, influence another variable, like output.

Output Function

An output function describes the quantity of goods or services produced (output) based on varying levels of input, like labor. In our example, the output function is \(Q(L) = 600L^{2/3}\), where \(Q\) depends on \(L\), the labor force size. Output functions are essential for businesses to understand the relationship between input resources and production output. They help in making informed decisions about resource allocation to achieve desired production levels. In our scenario, understanding the output function allows us to estimate how much labor needs to change to achieve a certain increase in output.

Marginal Analysis

Marginal analysis involves evaluating the additional benefits and costs of a decision. In economics, it often refers to the additional output obtained by increasing an input slightly. Using the derivative we calculated earlier (\(\frac{dQ}{dL} = 400L^{-1/3}\)), we can determine the marginal product of labor, which shows how much extra output is generated by employing an additional unit of labor. This analysis helps businesses optimize their input levels. In our exercise, marginal analysis is used to figure out the additional labor needed to achieve a 1% increase in output.

Percentage Change

Percentage change is a way of expressing how much a quantity increases or decreases in comparison to its original value. The manufacturer in the problem wants to increase the output by 1%. Using the output function and its derivative, we approximate the percentage increase in labor required. Given the desired output change as 1% \( (0.01 \times Q(L))\), we found that the labor change \(dL\) needed is \(\frac{3}{200} L\). Converting this to a percentage, we get 1.5%. Hence, to achieve a 1% increase in output, a 1.5% increase in labor is required. This concept is vital for understanding the efficiency and cost implications of scaling production.