Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider the Cobb-Douglas production model for a manufacturing process depending on three inputs \(x, y\), and \(z\) with unit costs \(a, b\), and \(c\), respectively, given by $$ P=k x^{\alpha} y^{\beta} z^{\gamma}, \quad \alpha>0, \beta>0, \gamma>0, \alpha+\beta+\gamma=1 $$ subject to the cost constraint \(a x+b y+c z=d\). Determine \(x, y\), and \(z\) to maximize the production \(P\).

Step-by-Step Solution

Answer

Maximize by setting \( x = \frac{\alpha d}{a} \), \( y = \frac{\beta d}{b} \), and \( z = \frac{\gamma d}{c} \).

Step 1: Set up the Lagrangian

To maximize the production function subject to the cost constraint, we use the method of Lagrange multipliers. The Lagrangian function is given by:\[ \mathcal{L} = k x^{\alpha} y^{\beta} z^{\gamma} + \lambda (d - a x - b y - c z) \] where \( \lambda \) is the Lagrange multiplier.

Step 2: Compute the Partial Derivatives

Take the partial derivatives of the Lagrangian with respect to each variable \(x, y, z\), and the Lagrange multiplier \( \lambda \):- \( \frac{\partial \mathcal{L}}{\partial x} = \alpha k x^{\alpha-1} y^{\beta} z^{\gamma} - \lambda a \)- \( \frac{\partial \mathcal{L}}{\partial y} = \beta k x^{\alpha} y^{\beta-1} z^{\gamma} - \lambda b \)- \( \frac{\partial \mathcal{L}}{\partial z} = \gamma k x^{\alpha} y^{\beta} z^{\gamma-1} - \lambda c \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = d - ax - by - cz \)

Step 3: Set Equations to Zero

To find the critical points, set all the partial derivatives of the Lagrangian equal to zero:\[ \alpha k x^{\alpha-1} y^{\beta} z^{\gamma} = \lambda a \]\[ \beta k x^{\alpha} y^{\beta-1} z^{\gamma} = \lambda b \]\[ \gamma k x^{\alpha} y^{\beta} z^{\gamma-1} = \lambda c \]\[ d = ax + by + cz \]

Step 4: Solve the System of Equations

Divide the first three equations by each other to eliminate \( \lambda \):1. \( \frac{\alpha}{a} = \frac{y}{x} \frac{\beta}{b} \)2. \( \frac{\beta}{b} = \frac{z}{y} \frac{\gamma}{c} \)3. \( \frac{\gamma}{c} = \frac{x}{z} \frac{\alpha}{a} \)This gives relationships between ratios of \(x, y, z\). Use these relations to express \(y\) and \(z\) in terms of \(x\). Finally, solve for \(x, y, z\) using the cost constraint.

Step 5: Express Each Variable

From equations obtained in Step 4, express variables as:\[ x = \frac{\alpha d}{a} \]\[ y = \frac{\beta d}{b} \]\[ z = \frac{\gamma d}{c} \]These expressions respect the ratios dictated by the coefficients in the production function and satisfy the cost constraint.

Key Concepts

Cobb-Douglas production function

The Cobb-Douglas production function is an economic model used to describe the relationship between multiple inputs and the output they produce. In our exercise, we deal with three inputs: \(x\), \(y\), and \(z\). These might represent resources like labor, capital, and raw materials. The function is expressed as \(P = k x^{\alpha} y^{\beta} z^{\gamma}\), where \(\alpha\), \(\beta\), and \(\gamma\) are positive constants, each representing the output elasticity of the respective input. This means each constant indicates the responsiveness of output \(P\) when the respective input changes, assuming others are constant. Another key idea here is that the sum of these constants, \(\alpha + \beta + \gamma = 1\), reflects constant returns to scale. This denotes that scaling up all inputs by any factor will proportionately scale up the output. Understanding this framework is crucial as it sets the foundation for relating changes in inputs to changes in output efficiently.

Partial derivatives

Partial derivatives are a mathematical tool that helps us understand how a function changes as one of its input variables changes, while all others are held constant. In the context of optimizing the Cobb-Douglas production function, partial derivatives are used to determine how the output \(P\) changes with small changes in \(x\), \(y\), and \(z\). Here's how we use them in our exercise: we took the partial derivative of our Lagrangian function \(\mathcal{L}\) with respect to each variable \(x\), \(y\), and \(z\). These derivatives tell us how sensitive the production is to each input when others are fixed. By setting these derivatives to zero, we find the stationary points, which are potential candidates for maximum or minimum values.

• For \(x\), we get: \(\frac{\partial \mathcal{L}}{\partial x} = \alpha k x^{\alpha-1} y^{\beta} z^{\gamma} - \lambda a\).
• The same approach is applied for \(y\) and \(z\).

These equations provide conditions that need to be true for the maximum production under given constraints.

Cost constraint

A cost constraint in this context restricts the total cost of input resources to a fixed budget. It is outlined as \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are the unit costs for inputs \(x\), \(y\), and \(z\) respectively, and \(d\) is the maximum cost allowable. This constraint is essential because it ensures our solution doesn't just identify maximum production blindly; instead, it respects the real-world limitation of budget.Using this constraint, you align the spending on inputs with the budget available, ensuring practicality. As part of the optimization process, this constraint is incorporated into the Lagrangian function. This combination helps ascertain how to allocate budget efficiently across \(x\), \(y\), and \(z\) to keep the cost within the limit yet achieve optimal output.

Optimization in calculus

Optimization in calculus involves finding values of variables that maximize or minimize a function, given certain conditions. In our case, you want to maximize production \(P\) using the Cobb-Douglas function under a specified cost constraint.To do this, we employ the method of Lagrange multipliers. This approach involves setting up a Lagrangian function \(\mathcal{L}\), combining our production function and the cost constraint, and then solving for the variables where the system achieves maximum production.

• First, we find the critical points by setting the partial derivatives equal to zero.
• Solving these equations alongside the cost constraint tells us how each input should be adjusted.

Every step ensures that the achievable maximum output is aligned with realistic economic conditions. This form of optimization shows how theoretical calculus tools can solve practical economic problems efficiently.