Economics Terms A-Z - The most important terms in economics.

a b c d e f g h i j k l m n o p q r s t u v w x y z

Economics Terms A-Z

Lagrangian Optimization

Read a summary or generate practice questions using the INOMICS AI tool

By , reviewed by Tom McKenzie

Lagrangian optimization is a method for solving optimization problems with constraints. The method makes use of the Lagrange multiplier, which is what gives it its name (this, in turn, being named after mathematician and astronomer Joseph-Louis Lagrange, born 1736). Many subfields of economics use this technique, and it is covered in most introductory microeconomics courses, so it pays to become familiar with it.

Often, microeconomists are interested in the optimal choice that an economic agent can make using their limited budget. For instance, what is the maximum utility someone can achieve by purchasing some combination of goods?

Lagrangian optimization allows us to solve this type of problem in a rather straightforward way. First, it assumes that a budget constraint is binding. This makes sense in a world where we assume the actors are rational consumers. If someone did not use all of their money, they understand that they would become better off by using the rest of their money.

A budget constraint will often look similar to q*p â‰¤ a, where a is the total budget, q is quantity, and p is price. For our problem, we will use the constraint q1*p1 + q2*p2 â‰¤ a, where q1 and q2 are two different goods each with their own prices. This shows that the total value of goods the consumer purchases must be less than or equal to the money they have. When a budget constraint like this is “binding”, that just means that the inequality becomes an equals sign. Or, in words, the consumer uses all of their money.

The next step is to subtract so that the budget constraint is all on the same side of the equation, like so: q1*p1 + q2*p2 - a = 0.

Now, we multiply this budget constraint by the Lagrange multiplier λ, which represents the “shadow price”. That is, when this problem is solved, the Lagrange multiplier λ shows us how much the objective function (i.e., the utility function, revenue function, etc.) changes in response to an infinitesimal change in the budget constraint.

To finish setting up the problem, simply subtract this modified budget constraint from the original equation. In our case, this means the full Lagrangian optimization problem looks like this:

L(q1,q2,p1,p2) = U(q1,q2,p1,p2) - λ(q1*p1 + q2*p2 - a)

where L denotes a Lagrangian optimization problem, and U(q1,q2,p1,p2) is the utility function as a function of q and p for both goods.

To solve this problem, we take a series of partial derivatives for each variable and set each equal to zero to find the critical points of this function. In this case, we assume that prices and the budget are fixed, which leaves q1, q2 and λ as variables that we should derive the Lagrangian with respect to. This will look like so:

        \frac{\partial{L}}{\partial{q}_1} = \frac{\partial{U}}{\partial{q}_1} - \lambda*\mathit{p}_1 = 0
        \frac{\partial{L}}{\partial{q}_2} = \frac{\partial{U}}{\partial{q}_2} - \lambda*\mathit{p}_2 = 0
        \frac{\partial{L}}{\partial{\lambda}} = 0 - \mathit{q}_1*\mathit{p}_1 - \mathit{q}_2*\mathit{p}_2 + \textit{a} = 0

Note that the last partial derivative simplifies to exactly the budget constraint we originally had. These partial derivatives give us a series of equations to solve.

In the real world, computer software can help us do this, but in economics courses the relations will be solvable via algebra. The solution yields the optimal amount of both goods for the consumer to purchase to maximize their utility, given their budget and the prices of both goods. These answers can be plugged back into the original utility function to yield the maximum amount of utility this provides.

As an example, consider that good 1 is tea (measured in liters), and good 2 are biscuits. Also suppose that the final results of the optimization problem give us q1 = 5 and q2 = 10. These amounts show that for this individual and their budget constraint, they achieve the highest level of utility possible by consuming 5 liters of tea and 10 biscuits. Then, q1 = 5 and q2 = 10 can be inserted back into U(q1,q2,p1,p2) along with the given prices to yield the total utility from this consumption mix for this individual. At this point, λ shows the additional utility this individual could gain if their budget increased by one dollar.

As a side note, you may be wondering how savings fits into this picture. People are often encouraged to keep some money set aside to save and not spend. How then can it be best for people to spend all of their money, so that the budget constraint is entirely used up?

In our framework with the budget constraint in this problem, “savings” can be thought of as a good just like a banana or a scarf. People choose how much money to “spend” on savings. Thus, the framework of our problem does not change. People are better off making use of their entire budget, where some part of the budget can be used to “purchase” savings. In macroeconomics, savings is usually considered separately from consumption as it is studied more closely than in our simple example here.

Good to Know

This type of problem can be easily adjusted to account for many different economic scenarios without changing the basic structure of the solution. For example, instead of considering a consumer’s utility function and their budget, we can analyze production decisions for a firm.

To do so, we replace the utility function with a revenue function, and the budget becomes the quantity times the cost for each good, rather than the price. The underlying math is not changed. When we plug the optimal amounts into the original revenue function, this yields the total possible revenue the firm can generate given their costs.

More advanced microeconomics courses will often return to this type of optimization with minor additions. For example, Bertrand, Cournot, and Stackelberg competition models can be solved using the Lagrange method with slight adjustments to the above formula.

Practice using the Lagrange method...

Exercises for you

You need to login to comment


The INOMICS AI can generate an article summary or practice questions related to the content of this article.
Try it now!

An error occured

Please try again later.

3 Practical questions, generated by our AI model