Question

Consider the following constrained maximization problem: \\[ \begin{array}{ll} \text { maximize } & y=x_{1}+5 \ln x_{2} \\ \text { subject to } & k-x_{1}-x_{2}=0 \end{array} \\] where \(k\) is a constant that can be assigned any specific value. a. Show that if \(k=10\), this problem can be solved as one involving only equality constraints. b. Show that solving this problem for \(k=4\) requires that \(x_{1}=-1\) c. If the \(x^{\prime}\) s in this problem must be non-negative, what is the optimal solution when \(k=4 ?\) (This problem may be solved either intuitively or using the methods outlined in the chapter.) d. What is the solution for this problem when \(k=20 ?\) What do you conclude by comparing this solution with the solution for part (a)? Note: This problem involves what is called a quasi-linear function. Such functions provide important examples of some types of behavior in consumer theory-as we shall see.

Short answer

Based on the provided step-by-step solution, provide a short answer discussing the results for the different values of k: When k = 10, the optimal solution is x1 = 5 and x2 = 5. When k = 4, the optimal solution cannot be found as there's no feasible solution, since x1 must be equal to -1, which is not possible under non-negativity constraint. When k = 20, the optimal solution is x1 = 15 and x2 = 5. From these results, it is observed that increasing the value of k causes an increase in the optimal value of x1 while leaving the optimal value of x2 unchanged, which indicates that the constraint is relatively more binding on x2, and x1 responds to the changes in k.

Step-by-step solution

Write down the Lagrangian function

To begin with, let's write down the Lagrangian function: \\[ L(x_{1}, x_{2}, \lambda) = x_{1} + 5 \ln{x_{2}} + \lambda (k - x_{1} - x_{2}) \\] The Lagrangian function L is the objective function plus the constraint multiplied by a Lagrange multiplier λ.

Find the first order conditions (FOC)

Now, we will differentiate the Lagrangian function with respect to x1 and x2 to obtain the FOC equations: \\[ \frac{\partial L(x_{1}, x_{2}, \lambda)}{\partial x_{1}} = 1 - \lambda \\] \\[ \frac{\partial L(x_{1}, x_{2}, \lambda)}{\partial x_{2}} = \frac{5}{x_{2}} - \lambda \\]

Solve the FOC equations and the constraint

We'll use the constraint equation k - x1 - x2 = 0 to eliminate x2 and substitute it into the second FOC equation: \\[ x_{2} = k - x_{1} \\] Now, substitute this into the second FOC equation: \\[ \frac{5}{k - x_{1}} - \lambda = 0 \\] At this point, we see that the two FOC equations are λ = 1 and λ = 5/(k-x1). We can equate these two expressions for λ and solve for x1: \\[ 1 = \frac{5}{k - x_{1}} \\] \\[ x_{1} = k - 5 \\] Now, we can plug this back into the constraint equation to find the optimal value for x2: \\[ k - 5 - x_{2} = 0 \\] \\[ x_{2} = k - 5 \\] So, the optimal solution is x1 = k - 5 and x2 = 5.

Find the optimal values of x1 and x2 for different values of k and discuss the results

Now, let's find the optimal values of x1 and x2 for k = 10, 4, and 20: a. For k = 10, x1 = 5 and x2 = 5. This problem can be solved as one involving only equality constraints. b. For k = 4, x1 = -1 and x2 = 5. To satisfy the constraint k = 4, x1 must be equal to -1. c. When x's are non-negative, the optimal solution for k = 4 can't be found because there is no feasible solution. One possible remedy would be to impose an additional constraint specifying non-negativity. d. For k = 20, x1 = 15 and x2 = 5. By comparing this solution with the solution when k = 10, we can observe that increasing the value of k causes an increase in the optimal value of x1 while leaving the optimal value of x2 unchanged. This tells us that the constraint is relatively more binding on x2, and x1 responds to the changes in k.

Key concepts

Lagrangian Function

At the heart of many economic optimization problems lies the Lagrangian function, a fundamental tool in constrained optimization. It allows us to transform a constrained problem into an unconstrained one by introducing a new variable, the Lagrange multiplier. To construct the Lagrangian function, we combine our objective function—in this case, maximizing output, given by \( y=x_{1}+5 \ln x_{2} \)—with the constraint \( k-x_{1}-x_{2}=0 \), multiplied by the Lagrange multiplier, \( \lambda \).

The resulting Lagrangian function for our exercise is:
\[ L(x_{1}, x_{2}, \lambda) = x_{1} + 5 \ln{x_{2}} + \lambda (k - x_{1} - x_{2}) \]
This function is then used to find the values of \( x_1 \), \( x_2 \), and \( \lambda \) that maximize the original function, while respecting the given constraint.

First Order Conditions (FOC)

The First Order Conditions, or FOCs, are critical to finding the optimum values in a constrained optimization problem. These conditions require setting the partial derivatives of the Lagrangian function with respect to each variable equal to zero.

For our problem, we calculate the derivatives with respect to \( x_1 \) and \( x_2 \):
\[ \frac{\partial L(x_{1}, x_{2}, \lambda)}{\partial x_{1}} = 1 - \lambda \]
\[ \frac{\partial L(x_{1}, x_{2}, \lambda)}{\partial x_{2}} = \frac{5}{x_{2}} - \lambda \]
Setting these derivatives equal to zero allows us to solve for the optimal values that satisfy both the objective function and the constraints. Solving the FOC equations alongside the constraint gives us the solution space for our optimization problem.

Quasi-linear Function

A quasi-linear function plays an important role in economics, especially in consumer theory. These functions typically have one linear term and one non-linear term. In our exercise, the objective function \( y=x_{1}+5 \ln x_{2} \) is quasi-linear because it has a linear term in \( x_1 \) and a non-linear logarithmic term in \( x_2 \).

Quasi-linear functions illustrate how consumers might make trade-offs between goods when one good can be consumed in large quantities without much effect on utility (the linear term), while increases in the other good lead to diminishing marginal returns to utility (the logarithmic term). They are particularly useful for analyzing consumer behavior under different types of market conditions and income levels.

Consumer Theory

Consumer theory examines how individuals decide to spend their income on the consumption of goods and services. It involves concepts like utilities, preferences, budget constraints, and consumption bundles.

In our exercise, the constrained maximization problem can be viewed through the lens of consumer theory, where \( x_1 \) and \( x_2 \) represent two different goods, and \( k \) is akin to the consumer's income or wealth. Consumers must decide the optimal bundle of \( x_1 \) and \( x_2 \) that provides the maximum utility, symbolized by \( y \), within their budget constraint \( k \). When determining solutions for different values of \( k \), we effectively analyze how changes in income affect consumption choices - a critical aspect of consumer theory.

Lagrange Multiplier

The Lagrange multiplier, \( \lambda \), is more than just an additional variable in the Lagrangian function; it holds significant economic interpretation. It represents the change in the maximum value of the objective function for a one-unit increase in the constraint bound.

In our example, observing how the multiplier changes as we solve for different values of \( k \) can reveal how sensitive our optimal solution is to the constraint. When \( k \) changes, the value of the multiplier could provide insight into how restrictive the budget constraint is on the consumer's utility maximization problem. The Lagrange multiplier plays a critical role not only in solving the constrained optimization but also in understanding the underlying economic principles of the problem.