Applying the Lagrangian Method to Consumer Theory

Constrained Optimization Basic Math for Economics – Refresher 1 Introduction  With an introduction to constraints out of the way, we are ready to talk about constrained optimization.  Recall that in these situations, we are trying to optimize some objective function subject to one or more constraints. These constraints may not even be binding, so they must be checked. Otherwise, we make use of the Lagrangian technique and use our first and second-order conditions as before. Let’s look at a couple of examples. 2 Constrained Optimization  Consider a situation where a consumer is deciding how much of goods 𝑥 and 𝑦 to purchase and receives utility from these two goods as follows: 𝑢 𝑥, 𝑦 = 𝑥 0.5 𝑦 0.5  In addition, the consumer has an income level of 𝐼 and faces prices 𝑝𝑥 and 𝑝𝑦 for goods 𝑥 and 𝑦, respectively.  Find the consumer’s optimal bundle of goods 𝑥 and 𝑦 as a function of income 𝐼, and the prices of each good. Let’s set this optimization problem up. 3 Constrained Optimization  This is a utility maximization problem for choice variables 𝑥 and 𝑦. In addition, our objective function is the utility function.  Adding all of this to our optimization problem, we have max 𝑥 0.5 𝑦 0.5 𝑥,𝑦  We also need to consider our budget constraint.  The total amount that our consumer spends on goods 𝑥 and 𝑦 cannot exceed their income, 𝐼, i.e., 𝑝𝑥 𝑥 + 𝑝𝑦 𝑦 ≤ 𝐼 4 Constrained Optimization 𝑝𝑥 𝑥 + 𝑝𝑦 𝑦 ≤ 𝐼  Rearranging our constraint such that it is greater than or equal to zero, 𝐼 − 𝑝𝑥 𝑥 − 𝑝𝑦 𝑦 ≥ 0  Now we assemble our Lagrangian by inserting the constraint along with our objective function (don’t forget to include a Lagrange multiplier). max 𝑥 0.5 𝑦 0.5 + 𝜆 𝐼 − 𝑝𝑥 𝑥 − 𝑝𝑦 𝑦 𝑥,𝑦,𝜆 We are now ready to calculate first-order conditions. 5 Constrained Optimization max 𝑥 0.5 𝑦 0.5 + 𝜆 𝐼 − 𝑝𝑥 𝑥 − 𝑝𝑦 𝑦 𝑥,𝑦,𝜆  With three choice variables, we need three first-order conditions.  Taking the derivative of our Lagrangian for each of our choice variables and setting them equal to zero yields, 𝜕ℒ = 0.5𝑥 −0.5 𝑦 0.5 − 𝜆𝑝𝑥 = 0 𝜕𝑥 𝜕ℒ = 0.5𝑥 0.5 𝑦 −0.5 − 𝜆𝑝𝑦 = 0 𝜕𝑦 𝜕ℒ = 𝐼 − 𝑝𝑥 𝑥 − 𝑝𝑦 𝑦 = 0 𝜕𝜆  This is just three equations and three unknowns, which we can solve. 6 Constrained Optimization 𝜕ℒ = 0.5𝑥 −0.5 𝑦 0.5 − 𝜆𝑝𝑥 = 0 𝜕𝑥 𝜕ℒ = 0.5𝑥 0.5 𝑦 −0.5 − 𝜆𝑝𝑦 = 0 𝜕𝑦 𝜕ℒ = 𝐼 − 𝑝𝑥 𝑥 − 𝑝𝑦 𝑦 = 0 𝜕𝜆  Let’s focus on the first two equations. Rearranging them, we have 0.5𝑥 −0.5 𝑦 0.5 = 𝜆𝑝𝑥 0.5𝑥 0.5 𝑦 −0.5 = 𝜆𝑝𝑦 7 Constrained Optimization 0.5𝑥 −0.5 𝑦 0.5 = 𝜆𝑝𝑥 0.5𝑥 0.5 𝑦 −0.5 = 𝜆𝑝𝑦  Next, we divide the first equation by the second, which cancels out 𝜆. 0.5𝑥 −0.5 𝑦 0.5 𝜆𝑝𝑋 𝑝𝑥 = = 0.5 −0.5 0.5𝑥 𝑦 𝜆𝑝𝑦 𝑝𝑦  We can then simplify this expression to obtain 𝑦 𝑝𝑥 = 𝑥 𝑝𝑦 𝑝𝑋 𝑥 = 𝑝𝑦 𝑦  This is known as a tangency condition. 8 Constrained Optimization 𝑝𝑋 𝑥 = 𝑝𝑦 𝑦 𝐼 − 𝑝𝑥 𝑥 − 𝑝𝑦 𝑦 = 0  Now we use the tangency condition along with our final first-order condition. Substituting the tangency condition in, 𝐼 − 𝑝𝑥 𝑥 − 𝑝 ต 𝑥𝑥 = 0 𝑝𝑦 𝑦  Rearranging terms, we have 2𝑝𝑥 𝑥 = 𝐼 9 Constrained Optimization   2𝑝𝑥 𝑥 = 𝐼 Next, we divide both sides by 2𝑝𝑥 to obtain our equilibrium value of good 𝑥, 𝐼 ∗ 𝑥 = 2𝑝𝑥 From here, we can use the tangency condition to solve for our equilibrium value of good 𝑦, 𝐼 ∗ 𝑝𝑦 𝑦 = 𝑝𝑥 𝑥 = 𝑝𝑥 2𝑝𝑥 𝐼 ∗ 𝑦 = 2𝑝𝑦  And we have our equilibrium bundle! 10 Constrained Optimization  Lastly, we need to check the value of 𝜆 to ensure that the constraint binds.  This can be done with either of the first two first-order conditions. Using the first one, 0.5𝑥 −0.5 𝑦 0.5 − 𝜆𝑝𝑥 = 0  Rearranging terms and plugging in our equilibria, 1 𝐼 𝜆= 2𝑝𝑥 2𝑝𝑥 −0.5 𝐼 2𝑝𝑦 0.5 1 = 2𝑝𝑥 𝑝𝑦 0.5 Since prices are positive, we can assume that 𝜆 > 0 and our constraint binds. 11 Constrained Optimization  We can continue and check our second-order conditions as well. I will leave that for you to try on your own. You’ll find that we do, in fact, have a maximum.  Let’s look at one more example, with a different utility function. This time, the consumer receive utility from goods 𝑥 and 𝑦 of 𝑢 𝑥, 𝑦 = − 𝑥 − 2 2 𝑦 − 3 2 In addition, the consumer has an income level of 𝐼 = 11, and faces prices of 𝑝𝑥 = 𝑝𝑦 = 1. 12 Constrained Optimization  Once again, we can set up their Lagrangian by adding their constraint to their optimization problem to obtain, max − 𝑥 − 2 2 𝑦 − 3 2 + 𝜆(11 − 𝑥 − 𝑦) 𝑥,𝑦,𝜆  Calculating first-order conditions, 𝜕ℒ = −2 𝑥 − 2 𝑦 − 3 2 − 𝜆 = 0 𝜕𝑥 𝜕ℒ = −2 𝑥 − 2 2 (𝑦 − 3) − 𝜆 = 0 𝜕𝑦 𝜕ℒ = 11 − 𝑥 − 𝑦 = 0 𝜕𝜆  Once again, this is just three equations and three unknowns. 13 Constrained Optimization 𝜕ℒ = −2 𝑥 − 2 𝑦 − 3 2 − 𝜆 = 0 𝜕𝑥 𝜕ℒ = −2 𝑥 − 2 2 (𝑦 − 3) − 𝜆 = 0 𝜕𝑦  Rearranging and combining the first two equations, we have, −2 𝑥 − 2 𝑦 − 3 2 𝜆 = =1 2 −2 𝑥 − 2 (𝑦 − 3) 𝜆  Rearranging once again, we obtain our tangency condition, 𝑦−3 =1 𝑥−2 𝑦 =𝑥+1 14 Constrained Optimization 𝑦 =𝑥+1 𝜕ℒ = 11 − 𝑥 − 𝑦 = 0 𝜕𝜆  Substituting our tangency condition into our third firstorder condition, 11 − 𝑥 − (𝑥 + 1)= 0 𝑦  Solving this expression for 𝑥, we obtain our equilibrium quantity, 𝑥∗ = 5 15 Constrained Optimization  From our tangency condition we obtain our equilibrium quantity of 𝑦, 𝑦∗ = 𝑥∗ + 1 = 6  Finally, we check to see if our constraint binds. Using the first first-order condition and solving it for 𝜆, 𝜆 = −2 𝑥 − 2 𝑦 − 3 2 = −2 5 − 2 6 − 3 2 = −54  This time the Lagrange multiplier comes out negative!  This implies that our constrain does not bind, and is in fact pushing our equilibrium away from its maximum point. Thus, our equilibrium is not utility maximizing. 16 Constrained Optimization  To fix this, we simply throw out the constraint and treat this problem as an unconstrained optimization problem.  Work through it, and you’ll find that the consumer optimizes by selecting 𝑥 ∗ = 2 and 𝑦 ∗ = 3. In this special kind of utility function, the consumer has a bliss point at exactly this level of consumption. They would actually rather keep some of their money even though they could purchase more. This is why it is important to check the values of the Lagrange multiplier. 17