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## Multivariable calculus

### Course: Multivariable calculus>Unit 3

Lesson 5: Lagrange multipliers and constrained optimization

# Meaning of the Lagrange multiplier

In the previous videos on Lagrange multipliers, the Lagrange multiplier itself has just been some proportionality constant that we didn't care about. Here, you can see what its real meaning is. Created by Grant Sanderson.

## Want to join the conversation?

• Will there always be only one value for lambda? • Each set of solutions will have one lambda. That is, if you are trying to find extrema for f(x,y) under the constraint g(x,y) = b, you will get a set of points (x1,y1), (x2,y2), etc that represent local mins and maxs. At each of these, there will be a single lambda. There is no guarantee that all the lambda will be the same (it is quite likely they will differ from each other).
• The insight at the end of the video is that the max lambda indicates the revenue per dollar. A lambda of 2.3 indicates that for every 1\$ invested the revenue will be 2.3\$. Does this hold only for small changes in the constraint as we are dealing with differential calculus (i.e. changes of a few percent in the budget constraint) or does this insight hold also for very large changes in the constraint (using the example in the video, changing the budget from 10,000 to 100,000)? • This interpretation of the Lagrange Multiplier (where lambda is some constant, such as 2.3) strictly holds only for an infinitesimally small change in the constraint. It will probably be a very good estimate as you make small finite changes, and will likely be a poor estimate as you make large changes in the constraint.
That being said, if you could find lambda as a function of b (this is easier said than done), you could integrate over your increase in budget to find the increase in revenue. This would hold true for any increase in budget, no matter how large.
• So h* for example, is dependent on b because when b, the constraint, changes, then the critical values also change, correct? • Since Grant specifies that this function is being maximized under a certain constrained function, I was wondering if there's a separate procedure for handling optimizations where we need to find a minimum value within a constraint. For instance, if we want to find the input points that would give the minimum value for an error function within a certain constraint of values, would the procedure still remain the same, or would it change to handle optimizations for finding minima?
(1 vote) • It is very common to use optimization to minimize a function. Imagine, for example, that you want to minimize a cost or an error.

The procedure Grant has shown here finds both the inputs which maximize R and the inputs which minimize R. This happens because this method finds all points at which the gradient of R is proportional to the gradient of the constraint, and the gradients are proportional at both maximum and minimum points.

If you recall from previous videos, Grant worked through the problem to find multiple possible solutions. He then plugged each solution back into his Revenue function to find which ones yielded the highest revenue. It turns out that (at least) one of the solutions he found (and discarded) minimizes the Revenue function.

As a side note, it turns out that all optimization procedures which maximize a function can be used to minimize a function simply by putting a negative in front of the function you wish to minimize.
• My microeconomics textbook has positive signs for the lagrange multipliers.like L = f(x1,x2...xn) + lambda*g(x1,x2...xn).
is the sign of the lagrange multiplier irrelevant and we can use whichever we want?
(1 vote) • At , is it possible that R(h, s) will be maximum at less budget than B(h, s) = 10k. That is R(h1, s1) > R(h2, s2), where B(h1, s1) < 10k and B(h2, s2) = 10k. We are still finding values for (h, s).
(1 vote) • In general, YES!
We would, however, have to formulate our problem slightly differently. Our problem statement said that we would spend \$10,000. A better way to formulate the problem would be to say that we can spend up to \$10,000.
Optimizing with this new constraint (an inequality constraint) tends to be more difficult, but can still be done (by combining the techniques in this module with those presented in the previous module). With this adjusted constraint, we may find out that the highest revenue will be reached when we spend less than \$10,000. In this case, the budget would be what we call an inactive constraint.
If the Lagrange Multiplier were negative, it would be a sure sign that a higher revenue could be attained by spending less.  