Utility Maximization Using the Lagrangian

The first approach to finding the optimal consumption bundle is precisely the method we used in the chapter; the only difference is that we used calculus to derive the marginal utilities and then solved for the marginal rate of substitution. A second approach introduces something known as the Lagrange multiplier, or λ. The Lagrangian is a technique for transforming a constrained optimization problem into an unconstrained problem by combining the objective function and the constraint into one equation. λ is a variable that multiplies the constraint.

Suppose, for example, that the objective function is f(x,y) and the constraint is g(x,y) = 0. The Lagrangian equation is

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Now maximize the equation by taking the partial derivatives of the equation with respect to x, y, and λ, and set them equal to zero. Partial derivatives in this form are known as first-order conditions, or FOC:

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So, we have three equations and three unknowns and can solve the system of equations. Note that the third first-order condition is simply the constraint.

Let’s see how the Lagrangian can be applied to our consumer facing the utility-maximization problem:

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(Notice that we wrote the budget constraint so that it is equal to zero—this is important for how we set up the Lagrangian.) This equation can be rewritten in Lagrangian form as

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Take the first-order conditions (FOCs):

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Embedded in these three first-order conditions are the same three pieces of information we’ve seen before: the marginal utilities of X and Y, the goods’ prices, and the consumer’s income.

The Lagrange multiplier λ is in the first two first-order conditions. So, solve each of these equations for λ:

image

How can we interpret this Lagrange multiplier? First, recognize that the numerators are the marginal utilities of X and Y. In other words, at the optimum, image . Therefore, λ is the exchange rate between utility and income—an additional dollar of income allows the consumer to purchase additional goods that generate λ more units of utility. We can also see this in the Lagrangian: If income increases by $1, maximum utility increases by λ units. In other words, λ measures the marginal utility of income. For example, let’s say that λ is 0.5. Then, if you gain $1 more in income, you’ll gain 0.5 units of utility.

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Note that this expression for λ is the optimization condition in terms of the consumer’s marginal utility per dollar spent that we derived in the chapter. We can rearrange this to get exactly what we showed graphically in the text—that the marginal rate of substitution equals the ratio of the prices:

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We can then solve for (X*,Y*) exactly as we did in the first approach, starting by finding Y as a function of X using the equality from the first two conditions:

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Using the last first-order condition, we can plug this value for Y into the budget constraint:

IPXXPYY = 0

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figure it out 4A.1

Let’s revisit Figure It Out 4.4. Antonio gets utility from burgers (B) and fries (F) in the form

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His income is $20, the price of burgers is $5, and the price of fries is $2.

Find Antonio’s optimal consumption bundle.

Solution:

To find the optimal consumption bundle, we need to solve the consumer’s utility-maximization problem:

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The solution in the chapter uses the approach in which we solve for the MRSBF from the marginal utilities. If instead we use the Lagrangian, we begin by writing Antonio’s constrained optimization problem and then solve for the first-order conditions:

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FOC:

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Use the first two conditions to solve for λ:

0.5B–0.5F0.5 = 5λ

λ = 0.1B–0.5F0.5

0.5B0.5F–0.5 = 2λ

λ = 0.25B0.5F–0.5

Set the two expressions for λ equal to each other and solve for F as a function of B:

λ = 0.1B–0.5F0.5 = 0.25B0.5F–0.5

0.1F0.5F0.5 = 0.25B0.5B0.5

F = (10)0.25B = 2.5B

So, for every burger Antonio consumes at the optimum, he will consume 2.5 orders of fries. Substitute F = 2.5B into the third condition (the consumer’s budget constraint) and solve for the optimal bundle (B*,F*):

20 = 5B + 2F

20 = 5B + 2(2.5B)

20 = 10B

B* = 2 burgers

F* = 2.5B = 2.5(2) = 5 orders of fries

This is where we stopped when we solved Antonio’s constrained optimization problem using the first approach we presented in this appendix. But using the Lagrangian, we can also solve for one more variable: the marginal value of Antonio’s income, λ, when Antonio is maximizing his utility.

λ = 0.1B–0.5F0.5 = 0.1(2) 0.5(5)0.5 ≈ 0.16

Therefore, Antonio’s utility increases by 0.16 units of utility for every extra dollar of income he has.

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