Problem 22 - Applied Calculus

Question

The quantity, \(q\), of a product manufactured depends on the number of workers, \(W\), and the amount of capital invested, \(K\), and is given by the Cobb-Douglas function $$q=6 W^{3 / 4} K^{1 / 4}.$$ In addition, labor costs are $$\$ 10$$ per worker and capital costs are $$\$ 20$$ per unit and the budget is $$\$ 3000.$$ (a) What are the optimum number of workers and the optimum number of units of capital? (b) Recompute the optimum values of \(W\) and \(K\) when the budget is increased by $$\$ 1.$$ Check that increasing the budget by $$\$ 1$$ allows the production of \(\lambda\) extra units of the product, where \(\lambda\) is the Lagrange multiplier.

Step-by-Step Solution

Answer

Use Lagrange multipliers to solve, resulting in optimal \(W\) and \(K\), and verify \(\lambda\) as the production change rate with respect to budget.

Step 1: Understand the Budget Constraint

The budget constraint states that the total cost for labor and capital should not exceed the budget. Given labor costs \(\(10\) per worker and capital costs \(\)20\) per unit, the constraint can be expressed as:\[10W + 20K = 3000.\]

Step 2: Formulate the Lagrangian

To find the optimum values of \(W\) and \(K\), we use the method of Lagrange multipliers. The Lagrangian function is given by:\[\mathcal{L}(W, K, \lambda) = 6W^{3/4}K^{1/4} + \lambda(3000 - 10W - 20K).\]

Step 3: Take Partial Derivatives

Take the partial derivatives of the Lagrangian with respect to \(W\), \(K\), and \(\lambda\) and set them equal to zero:- For \(W\): \(\frac{\partial \mathcal{L}}{\partial W} = \frac{9}{2}W^{-1/4}K^{1/4} - 10\lambda = 0\)- For \(K\): \(\frac{\partial \mathcal{L}}{\partial K} = \frac{3}{2}W^{3/4}K^{-3/4} - 20\lambda = 0\)- For \(\lambda\): \(3000 - 10W - 20K = 0\)

Step 4: Solve the System of Equations

From the partial derivatives with respect to \(W\) and \(K\), we obtain:\[\frac{9}{2}W^{-1/4}K^{1/4} = 10\lambda\]\[\frac{3}{2}W^{3/4}K^{-3/4} = 20\lambda\]Solve these equations along with the budget constraint equation for \(W\), \(K\), and \(\lambda\).

Step 5: Recompute for Increased Budget

Increase the budget by 1 unit to 3001: \[10W + 20K = 3001.\]Repeat Steps 2 through 4 with the new budget to find the new values of \(W\), \(K\), and the change in quantity, verifying that \(\lambda\) gives the change in output.

Key Concepts

Lagrange multipliers

The method of Lagrange multipliers is a powerful tool for finding the maximum or minimum of a function subject to certain constraints. In optimization problems involving the Cobb-Douglas function, it is often used to address constraints such as budgets or resources. The idea is to extend your usual method of finding extrema using derivatives, to include a factor called a multiplier that considers the constraints.

In the context of our exercise, we introduce the Lagrange multiplier, \(\lambda\), which accounts for the budget constraint. We set up a new function, the Lagrangian, which combines both our function to optimize, \(6W^{3/4}K^{1/4}\), and the budget constraint, \(10W + 20K = 3000\), into one expression:
\[\mathcal{L}(W, K, \lambda) = 6W^{3/4}K^{1/4} + \lambda(3000 - 10W - 20K)\]

This method converts our constrained problem into an unconstrained one by introducing \(\lambda\). Solving this will help us determine not just the optimum values of \(W\) and \(K\), but also the impact of the budget constraint on the optimal solution.

budget constraint

A budget constraint represents the limitations on the expenditure for a given optimization problem, ensuring that the costs do not exceed a predefined budget. This is a critical concept in production and economics where the goal is often to maximize output or profit while staying within budget.

In our case, the budget constraint is expressed as:
\[10W + 20K = 3000\]
Here, \(10\) is the cost per worker, and \(20\) is the cost per unit of capital. The equation ensures that the combined costs don't surpass \(3000\). Such constraints make the problem realistic, mimicking real-world situations where resources are limited. The budget constraint is a linear equation, a common type of constraint in optimization problems like this.

partial derivatives

Partial derivatives are used to understand how a function changes as one of its variables changes, keeping the others constant. They're a fundamental tool in multivariable calculus, particularly useful in optimization problems.

When dealing with the Cobb-Douglas production function combined with a constraint, we'll take the partial derivatives of the Lagrangian with respect to each of its variables: \(W\), \(K\), and \(\lambda\). For example, the partial derivative with respect to \(W\) in our exercise would be:
\[\frac{\partial \mathcal{L}}{\partial W} = \frac{9}{2}W^{-1/4}K^{1/4} - 10\lambda\]

Setting this to zero, along with the derivatives with respect to \(K\) and \(\lambda\), helps find the critical points that give us the desired optimum values. This step effectively balances the trade-offs between different factors, accounting for how changing one factor impacts the production output.

optimization problem

An optimization problem involves finding the best possible solution from a set of available alternatives given certain criteria or constraints. In our exercise, it's about maximizing the quantity of output, \(q\), produced by determining the best possible combination of workers \(W\) and capital \(K\), while obeying the budget constraint.

We start by setting up the Lagrangian, taking partial derivatives, and solving the resulting system of equations. These steps help us identify the conditions under which the production is optimized. By incorporating the constraints through Lagrange multipliers, we're able to focus on the most feasible solution under given limitations, like budget constraints.

Such problems are common in economic and production scenarios, demonstrating how mathematical techniques can help businesses and economists make informed decisions, aligning objectives with resource limitations.