Question

The production function for a company is given by \(f(x, y)=100 x^{0.25} y^{0.75}\) where \(x\) is the number of units of labor and \(y\) is the number of units of capital. Suppose that labor costs \(\$ 48\) per unit and capital costs \(\$ 36\) per unit. The total cost of labor and capital is limited to \(\$ 100,000\). (a) Find the maximum production level for this manufacturer. (b) Find the marginal productivity of money. (c) Use the marginal productivity of money to find the maximum number of units that can be produced if \(\$ 125,000\) is available for labor and capital.

Short answer

The maximum production level for this manufacturer is approximately 14202 units. The marginal productivity of money is approximately 0.14202. If there's a total of \$125,000 available for labor and capital, the maximum number of units that can be produced is approximately 17753 units.

Step-by-step solution

Formulate the Lagrangian

The production function is given by \(f(x, y)=100 x^{0.25} y^{0.75}\), and the budget constraint is \(48x + 36y = 100,000\). Formulate the Lagrangian for the problem: \[\mathcal{L}(x, y, \lambda)= 100 x^{0.25} y^{0.75} - \lambda(48x + 36y - 100,000)\]

Find the Partial Derivatives

Take the partial derivatives of the Lagrangian with respect to x, y, and \(\lambda\) and set them equal to zero: \[\frac{\partial \mathcal{L}}{\partial x}= 25 x^{-0.75} y^{0.75} - 48\lambda = 0\] \[\frac{\partial \mathcal{L}}{\partial y}= 75 x^{0.25} y^{-0.25} - 36\lambda = 0\] \[\frac{\partial \mathcal{L}}{\partial \lambda}= 48x + 36y - 100,000 = 0\]

Solve the System of Equations

After solving the system of equations formed by the partial derivatives equals to zero, the optimum levels for labour and capital are obtained. This results in \(x \approx 578.0342\) and \(y \approx 961.4220\)

Find the Maximum Production Level

Substituting the optimum levels of labour and capital into the production function will give the maximum production level. By subtitution, \(f(578.0342,961.4220) \approx 14202.13\). This is the maximum production given the cost constraint

Find the Marginal Productivity

To find the marginal productivity of money, divide the maximum production by the given total cost (\$100,000). This is the additional production for each extra dollar spent. The marginal productivity is hence \(\frac{14202.13}{100,000} = 0.14202\)

Use the Marginal Productivity to find maximum number units of production for a different budget

If \$125,000 is available for labor and capital, multiply this amount by the marginal productivity found in the previous step. This results in \(0.14202 \times 125,000 = 17752.66\), indicating it can produce almost 17753 units of production given the increased budget.

Key concepts

Lagrangian Method

When faced with the challenge of optimizing a function—such as a production function—subject to certain constraints, one proven approach is utilizing the Lagrangian method. This mathematical technique introduces a new variable, often denoted as \(\lambda\), known as the Lagrange multiplier. It transforms the constraint optimization problem into an unconstrained problem that can be solved using calculus.

Conceptually, the Lagrangian \(\mathcal{L}\) for a production function combines the original function objectives, like maximizing output, and the constraints, like budget limitations. In essence, the method seeks to find the input quantities—in this case, labor and capital—that result in the highest production levels, while adhering to budget constraints.

The essential steps involve constructing the Lagrangian by adding the original function and the constraint modified by the multiplier. Follow this by calculating partial derivatives, setting them to zero to solve for the variables, which include input quantities and the Lagrange multiplier. This process eventually leads to finding the optimal solution.

Marginal Productivity

Right at the heart of production optimization is the concept of marginal productivity. It measures the additional output that can be produced by adding one more unit of a specific input, while keeping other inputs constant. In practical terms, it tells us the contribution of each additional unit of labor or capital to the total production.

Understanding marginal productivity is crucial for efficient resource allocation. If, for instance, the marginal productivity of an input is low, this indicates that increasing its quantity would yield little additional output. Conversely, a high marginal productivity implies that an input is effectively contributing to the production process and investment in additional units can significantly boost output.

In our optimization problem, after determining maximum production, marginal productivity of money calculates the efficiency of investment in terms of production growth per dollar—strikingly important for budget allocation and cost analysis.

Partial Derivatives

Optimization problems usually involve functions dependent on several variables. Partial derivatives become the cornerstone for analyzing such multivariable functions. They show how the function changes as one variable is varied, while the others are held constant.

For optimization with the Lagrangian method, one takes the partial derivatives of the function with respect to each variable—including the Lagrange multiplier—and the information gained from these derivatives is then used to understand and navigate the function’s behavior. The condition for optimality is achieved when these partial derivatives are equal to zero, implying a local extremum point of the function—hopefully the desired maximum production point in our case.

The precision offered by partial derivatives makes them indispensable for meticulously detailing the sensitivity of output in regards to variations in labor and capital, critical for making informed decisions in production management.

Budget Constraint

A budget constraint represents a real-life limitation faced by companies: there is only a finite amount of financial resources available for investments in inputs such as labor and capital. It is expressed as an equation that ties together the costs of these inputs and the total budget available.

In an optimization problem, the budget constraint is crucial as it defines the feasible set of input combinations that a company can afford. This constraint is then embedded into the Lagrangian function, which allows exploration within the bounds of financial viability. Thus, it characterizes the trade-offs a company must consider when allocating funds to various inputs, aiming to maximize the return on investment within the constraints of the budget.