Chapter 12: Problem 1
Given \(U=(x+2)(y+1)\) and \(P_{x}=4, P_{y}=6,\) and \(B=130\) (a) Write the Lagrangian function. (b) Find the optimal levels of purchase \(x^{*}\) and \(y^{*}\) (c) ts the second-order sufficient condition for maximum satisfied? (d) Does the answer in (b) give any comparative-static information?
Short Answer
Expert verified
(a) The Lagrangian: \( L = (x+2)(y+1) + \lambda(130 - 4x - 6y) \). (b) \(x^* = 16\), \(y^* = 11\). (c) Yes, the Hessian is negative definite. (d) Yes, it provides comparative-static insights.
Step by step solution
01
Introduction to the Problem
We have a utility function given by \( U = (x+2)(y+1) \) and need to optimize it subject to a budget constraint with prices \( P_x = 4 \) and \( P_y = 6 \) and a budget \( B = 130 \).
02
Write the Lagrangian Function
The Lagrangian function combines the utility function and the budget constraint. It is given by:\[ L(x, y, \lambda) = (x+2)(y+1) + \lambda (130 - 4x - 6y) \]where \(\lambda\) is the Lagrange multiplier representing the shadow price of income.
03
Determine the First-Order Conditions
To find the first-order conditions, take the partial derivatives of the Lagrangian with respect to \(x\), \(y\), and \(\lambda\) and set them equal to zero:\[ \frac{\partial L}{\partial x} = (y+1) - 4\lambda = 0 \] \[ \frac{\partial L}{\partial y} = (x+2) - 6\lambda = 0 \]\[ \frac{\partial L}{\partial \lambda} = 130 - 4x - 6y = 0 \]
04
Solve the Equations
From \( \frac{\partial L}{\partial x} = 0 \), we have \( \lambda = \frac{y+1}{4} \).From \( \frac{\partial L}{\partial y} = 0 \), \( \lambda = \frac{x+2}{6} \).Equating the two expressions for \( \lambda \), \( \frac{y+1}{4} = \frac{x+2}{6} \).Cross multiplying and simplifying gives \( 6(y+1) = 4(x+2) \).Solving: \( 6y + 6 = 4x + 8 \) simplifies to \( 4x = 6y - 2 \). Equation 4: \( x = \frac{3}{2}y - \frac{1}{2} \).Substitute \( x = \frac{3}{2}y - \frac{1}{2} \) into the budget constraint:\[ 4(\frac{3}{2}y - \frac{1}{2}) + 6y = 130 \].Simplifies to:\[ 6y = 132 \], hence \( y = 11 \) and \( x = \frac{3}{2}(11) - \frac{1}{2} = 16 \).
05
Second-Order Sufficient Conditions
The second-order conditions involve checking the determinant of the bordered Hessian matrix of the Lagrangian. We need condition: 1. The bordered Hessian should be negative definite.Calculate the bordered Hessian and evaluate:\[ H = \begin{vmatrix}0 & -4 & -6 \-4 & 0 & 0 \-6 & 0 & 0 \end{vmatrix} \]. Evaluate for negative definiteness ensuring optimal conditions are satisfied.
06
Comparative Static Analysis
Comparative-static analysis looks at how changes in parameters like income or prices affect optimal decisions. Here:- Changes in prices \(P_x\) or \(P_y\) or the budget \(B\) would shift the budget constraint.- Can be analyzed using the Lagrange multiplier \(\lambda\), representing the marginal utility of income in terms of utility gained per additional dollar spent.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lagrangian Function
In constrained optimization, the Lagrangian function is a key tool. It helps combine an objective function, in this case, the utility function, with constraints, like a budget or any other limitations you work with. The Lagrangian is given by:\[ L(x, y, \lambda) = (x+2)(y+1) + \lambda (130 - 4x - 6y) \]Here, \( \lambda \) is known as the Lagrange multiplier. It plays a critical role by representing the change in the objective function as a result of changes in the constraint.
- The term \((x+2)(y+1)\) represents the objective function, or utility in this case.
- The term \( \lambda (130 - 4x - 6y) \) ties the budget restriction to the utility maximization problem.
First-Order Conditions
To locate the optimum, we derive the first-order conditions (FOCs) from the Lagrangian. These conditions are essentially the derivatives of the Lagrangian with respect to each variable, which include the goods \(x\) and \(y\), and the constraint represented by \(\lambda\).For our function, these derivatives are:\[ \frac{\partial L}{\partial x} = (y+1) - 4\lambda = 0 \]\[ \frac{\partial L}{\partial y} = (x+2) - 6\lambda = 0 \]\[ \frac{\partial L}{\partial \lambda} = 130 - 4x - 6y = 0 \]
- First equation indicates the rate at which utility increases with each unit of \( x \) should counterbalance the reduction in the constraint.
- Same principle applies for \( y \) in the second equation.
- The final equation ensures satisfaction of the budget constraint.
Second-Order Sufficient Conditions
While first-order conditions identify candidate points for being optimum, the second-order sufficient conditions help confirm whether these points are indeed maxima. These involve checking the properties of the bordered Hessian matrix derived from the Lagrangian:The bordered Hessian is expressed as:\[ H = \begin{vmatrix} 0 & -4 & -6 \ -4 & 0 & 0 \ -6 & 0 & 0 \end{vmatrix} \]
- The determinant of the bordered Hessian must be negative-definite for a maximum.
- A negative-definite bounded Hessian implies the identified points not only meet the first-order conditions but are indeed the maxima, ensuring no better value for utility exists without breaching the budget constraint.
Comparative Statics
Comparative statics analyses how changes to constraints or parameters like prices and income influence optimal solutions. Here, the Lagrange multiplier \( \lambda \) serves as a key indicator of changes in optimization due to its ties to the budget constraint.Changes in:
- Income \( B \) will generally involve shifts in the budget constraint, affecting consumption choices \(x^*\) and \(y^*\).
- Price changes \(P_x\) or \(P_y\) alter the constraint and potentially the optimal purchase levels.