Problem 19 - Applied Calculus

Question

A company's production output, \(P\), is given in tons, and is a function of the number of workers, \(N\), and the value of the equipment, \(V\), in units of $$\$ 25,000.$$The production function for the company is $$ P=f(N, V)=5 N^{0.75} V^{0.25} $$ The company currently employs 80 workers, and has equipment worth $$\$ 750,000.$$What are \(N\) and \(V\) ? Find the values of \(f, f_{N}\), and \(f_{V}\) at these values of \(N\) and \(V\). Give units and explain what each answer means in terms of production.

Step-by-Step Solution

Answer

At \(N = 80\) and \(V = 30\), calculate \(P\), \(f_N\), and \(f_V\) to understand production and its sensitivities.

Step 1: Identify the Given Values

We are given that the company employs 80 workers. Hence, \(N = 80\). The equipment worth is \(\(750,000\). Since equipment value \(V\) is in units of \(\\)25,000\), we calculate \(V\) by dividing: \(V = \frac{750,000}{25,000} = 30\).

Step 2: Understand the Production Function

The production function is given by \(P = f(N, V) = 5 N^{0.75} V^{0.25}\). This function expresses the output in terms of the number of workers and the equipment value.

Step 3: Calculate Total Production

Substitute \(N = 80\) and \(V = 30\) into the production function: \(P = 5(80)^{0.75}(30)^{0.25}\). Calculate \(80^{0.75}\) and \(30^{0.25}\), and then multiply by 5 to find \(P\).

Step 4: Calculate Workers' Contribution to Production

The marginal product of labor, \(f_N\), is the partial derivative of \(f\) with respect to \(N\). Compute \(f_N(N, V) = \frac{\partial}{\partial N} (5 N^{0.75} V^{0.25}) = 0.75 \times 5 N^{-0.25} V^{0.25}\). Substitute \(N = 80\) and \(V = 30\) to find \(f_N\).

Step 5: Calculate Equipment's Contribution to Production

The marginal product of capital, \(f_V\), is the partial derivative of \(f\) with respect to \(V\). Compute \(f_V(N, V) = \frac{\partial}{\partial V} (5 N^{0.75} V^{0.25}) = 0.25 \times 5 N^{0.75} V^{-0.75}\). Substitute \(N = 80\) and \(V = 30\) to find \(f_V\).

Step 6: Summary of Results and Units Explanation

Summarize the numerical results: total production \(P\), the marginal contribution per additional worker \(f_N\) in tons per worker, and the marginal contribution per additional unit of equipment \(f_V\) in tons per unit of \(\$25,000\). These explain how production changes with changes in workers or equipment.

Key Concepts

Marginal Product of Labor

The marginal product of labor is a crucial concept when analyzing production functions. It tells us how much additional output, typically in this context measured in tons, is generated by adding one more worker to the production process.
In mathematical terms, it is expressed as the partial derivative of the production function with respect to labor (), denoted here as \(f_N\).
To visualize this, consider the production function given: \[P = 5 N^{0.75} V^{0.25}\]Taking the partial derivative with respect to \(N\), we obtain:\[f_N(N, V) = \frac{\partial}{\partial N} (5 N^{0.75} V^{0.25}) = 0.75 \times 5 N^{-0.25} V^{0.25}\]To compute the marginal product of labor for the company using 80 workers and equipment valued at \\(750,000 (or \(V=30\) in units of \\)25,000), substitute these values into the equation:
\[f_N(80, 30) = 0.75 \times 5 \times 80^{-0.25} \times 30^{0.25}\]Performing this calculation provides the exact marginal output per additional worker. The units are typically tons per worker, conveying the additional tons produced by employing one more laborer.

Marginal Product of Capital

Similar to labor, the marginal product of capital measures how much additional output is produced when one more unit of capital (in this case, equipment worth \\(25,000) is utilized. Its mathematical representation is the partial derivative of the production function with respect to capital (), symbolized as \(f_V\).
Let's revisit the production function:\[P = 5 N^{0.75} V^{0.25}\]By taking the partial derivative with respect to \(V\), we find the marginal product of capital:\[f_V(N, V) = \frac{\partial}{\partial V} (5 N^{0.75} V^{0.25}) = 0.25 \times 5 N^{0.75} V^{-0.75}\]For the given scenario (80 workers and equipment of \\)750,000), insert these into the equation:
\[f_V(80, 30) = 0.25 \times 5 \times 80^{0.75} \times 30^{-0.75}\]Solving this provides the increase in production due to an additional unit of equipment. The result is expressed in tons per unit of \$25,000, making it clear how much more is produced with each increment of capital investment.

Partial Derivatives

Partial derivatives play a significant role in understanding how different inputs affect the total output in a production function. When dealing with functions of multiple variables, like our production function with labor and capital, partial derivatives allow us to observe the change in output with respect to one variable while holding the others constant.
Suppose you have a production function given by:\[P = 5 N^{0.75} V^{0.25}\]Calculating the partial derivative with respect to \(N\) (labor) gives the marginal product of labor:\[f_N = \frac{\partial P}{\partial N} = 0.75 \times 5 N^{-0.25} V^{0.25}\]While computing the partial derivative with respect to \(V\) (capital) yields the marginal product of capital:\[f_V = \frac{\partial P}{\partial V} = 0.25 \times 5 N^{0.75} V^{-0.75}\]These calculations are essential for determining how changes to each respective input impact production. Partial derivatives, therefore, help managers make informed decisions about adding more workers or investing in more equipment to maximize efficiency and output.