Doing this gives,. This gave two possibilities. This leaves the second possibility. Therefore, the only solution that makes physical sense here is. We should be a little careful here. Anytime we get a single solution we really need to verify that it is a maximum or minimum if that is what we are looking for.
This is actually pretty simple to do. If the volume of this new set of dimensions is smaller that the volume above then we know that our solution does give a maximum. If, on the other hand, the new set of dimensions give a larger volume we have a problem. We only have a single solution and we know that a maximum exists and the method should generate that maximum. The only thing we need to worry about is that they will satisfy the constraint. So, we can freely pick two values and then use the constraint to determine the third value.
Plugging these into the constraint gives,. This is fairly standard for these kinds of problems. This one is going to be a little easier than the previous one since it only has two variables. To determine if we have maximums or minimums we just need to plug these into the function.
Also recall from the discussion at the start of this solution that we know these will be the minimum and maximums because the Extreme Value Theorem tells us that minimums and maximums will exist for this problem. Do not always expect this to happen. First note that our constraint is a sum of three positive or zero number and it must be 1. In each case two of the variables must be zero. We also have two possible cases to look at here as well.
However, this also means that,. So, we have four solutions that we need to check in the function to see whether we have minimums or maximums. We used it to make sure that we had a closed and bounded region to guarantee we would have absolute extrema.
For example. Before we proceed we need to address a quick issue that the last example illustrates about the method of Lagrange Multipliers.
We found the absolute minimum and maximum to the function. However, what we did not find is all the locations for the absolute minimum. Every point in this set of points will satisfy the constraint from the problem and in every case the function will evaluate to zero and so also give the absolute minimum. So, what is going on? Recall from the previous section that we had to check both the critical points and the boundaries to make sure we had the absolute extrema. The same was true in Calculus I.
We had to check both critical points and end points of the interval to make sure we had the absolute extrema. So, after going through the Lagrange Multiplier method we should then ask what happens at the end points of our variable ranges.
In the first three cases we get the points listed above that do happen to also give the absolute minimum. For the later three cases we can see that if one of the variables are 1 the other two must be zero to meet the constraint and those were actually found in the example.
In the case of this example the end points of each of the variable ranges gave absolute extrema but there is no reason to expect that to happen every time. The moral of this is that if we want to know that we have every location of the absolute extrema for a particular problem we should also check the end points of any variable ranges that we might have. If all we are interested in is the value of the absolute extrema then there is no reason to do this.
We can also have constraints that are inequalities. The main difference between the two types of problems is that we will also need to find all the critical points that satisfy the inequality in the constraint and check these in the function when we check the values we found using Lagrange Multipliers. Note that the constraint here is the inequality for the disk. Because this is a closed and bounded region the Extreme Value Theorem tells us that a minimum and maximum value must exist.
The first step is to find all the critical points that are in the disk i. This is easy enough to do for this problem. Here are the two first order partial derivatives.
At this point we proceed with Lagrange Multipliers and we treat the constraint as an equality instead of the inequality. We only need to deal with the inequality when finding the critical points. To find the maximum and minimum we need to simply plug these four points along with the critical point in the function. In this case, the minimum was interior to the disk and the maximum was on the boundary of the disk.
The final topic that we need to discuss in this section is what to do if we have more than one constraint. We will look only at two constraints, but we can naturally extend the work here to more than two constraints.
The system that we need to solve in this case is,. Connect and share knowledge within a single location that is structured and easy to search. When these point in the same direction, no possible movement makes any improvement.
I hope it's obvious that this is just a sloppy heap of intuition. Maybe it's a good exercise to figure out the qualifications that need to be made to make it closer to actually correct. I find lagrange multipliers easiest to understand in terms of vectors, so the concepts described apply mainly to functions of 2 variables but I imagine the concepts generalise readily to higher dimensions.
Similar considerations apply to maxima. Note that if there are multiple minima or maxima this should just fall out of the maths. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group.
Create a free Team What is Teams? Learn more. Why do Lagrange Multipliers work? Ask Question. Asked 6 years, 5 months ago. Active 3 years, 4 months ago. Viewed 3k times. Can someone please give me an explanation. Thank you! CivilSigma CivilSigma 2 2 gold badges 9 9 silver badges 22 22 bronze badges.
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