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August 7, 2018

Calculus III - Notes
Partial Derivatives Previous Chapter   Next Chapter Multiple Integrals
Absolute Minimums and Maximums Previous Section   Next Section Multiple Integrals (Introduction)

 Lagrange Multipliers

In the previous section we optimized (i.e. found the absolute extrema) a function on a region that contained its boundary.  Finding potential optimal points in the interior of the region isn’t too bad in general, all that we needed to do was find the critical points and plug them into the function.  However, as we saw in the examples finding potential optimal points on the boundary was often a fairly long and messy process. 


In this section we are going to take a look at another way of optimizing a function subject to given constraint(s).  The constraint(s) may be the equation(s) that describe the boundary of a region although in this section we won’t concentrate on those types of problems since this method just requires a general constraint and doesn’t really care where the constraint came from.


So, let’s get things set up.  We want to optimize (i.e. find the minimum and maximum value of) a function, , subject to the constraint .  Again, the constraint may be the equation that describes the boundary of a region or it may not be.  The process is actually fairly simple, although the work can still be a little overwhelming at times.


Method of Lagrange Multipliers

  1. Solve the following system of equations.



  1. Plug in all solutions, , from the first step into  and identify the minimum and maximum values, provided they exist.

The constant, , is called the Lagrange Multiplier.


Notice that the system of equations actually has four equations, we just wrote the system in a simpler form.  To see this let’s take the first equation and put in the definition of the gradient vector to see what we get.




In order for these two vectors to be equal the individual components must also be equal.  So, we actually have three equations here.



These three equations along with the constraint, , give four equations with four unknowns x, y, z, and .


Note as well that if we only have functions of two variables then we won’t have the third component of the gradient and so will only have three equations in three unknowns x, y, and .


As a final note we also need to be careful with the fact that in some cases minimums and maximums won’t exist even though the method will seem to imply that they do.  In every problem we’ll need to go back and make sure that our answers make sense.


Let’s work a couple of examples.


Example 1  Find the dimensions of the box with largest volume if the total surface area is 64 cm2.



Before we start the process here note that we also saw a way to solve this kind of problem in Calculus I, except in those problems we required a condition that related one of the sides of the box to the other sides so that we could get down to a volume and surface area function that only involved two variables.  We no longer need this condition for these problems.


Now, let’s get on to solving the problem.  We first need to identify the function that we’re going to optimize as well as the constraint.  Let’s set the length of the box to be x, the width of the box to be y and the height of the box to be z.  Let’s also note that because we’re dealing with the dimensions of a box it is safe to assume that x, y, and z are all positive quantities.


We want to find the largest volume and so the function that we want to optimize is given by,



Next we know that the surface area of the box must be a constant 64.  So this is the constraint.  The surface area of a box is simply the sum of the areas of each of the sides so the constraint is given by,



Note that we divided the constraint by 2 to simplify the equation a little.  Also, we get the function  from this.



Here are the four equations that we need to solve.






There are many ways to solve this system.  We’ll solve it in the following way.  Let’s multiply equation (1) by x, equation (2) by y and equation (3) by z.  This gives,





Now notice that we can set equations (5) and (6) equal.  Doing this gives,



This gave two possibilities.  The first,  is not possible since if this was the case equation (1) would reduce to


Since we are talking about the dimensions of a box neither of these are possible so we can discount .  This leaves the second possibility.



Since we know that  (again since we are talking about the dimensions of a box) we can cancel the z from both sides.  This gives,



Next, let’s set equations (6) and (7) equal.  Doing this gives,


As already discussed we know that  won’t work and so this leaves,


We can also say that  since we are dealing with the dimensions of a box so we must have,



Plugging equations (8) and (9) into equation (4) we get,




However, we know that y must be positive since we are talking about the dimensions of a box.  Therefore the only solution that makes physical sense here is




So, it looks like we’ve got a cube here.


We should be a little careful here.  Since we’ve only got one solution we might be tempted to assume that these are the dimensions that will give the largest volume.  The method of Lagrange Multipliers will give a set of points that will either maximize or minimize a given function subject to the constraint, provided there actually are minimums or maximums. 


The function itself,  will clearly have neither minimums or maximums unless we put some restrictions on the variables.  The only real restriction that we’ve got is that all the variables must be positive.  This, of course, instantly means that the function does have a minimum, zero.


The function will not have a maximum if all the variables are allowed to increase without bound.  That however, can’t happen because of the constraint,




Here we’ve got the sum of three positive numbers (because x, y, and z are positive) and the sum must equal 32.  So, if one of the variables gets very large, say x, then because each of the products must be less than 32 both y and z must be very small to make sure the first two terms are less than 32.  So, there is no way for all the variables to increase without bound and so it should make some sense that the function, , will have a maximum.

This isn’t a rigorous proof that the function will have a maximum, but it should help to visualize that in fact it should have a maximum and so we can say that we will get a maximum volume if the dimensions are : .


Notice that we never actually found values for  in the above example.  This is fairly standard for these kinds of problems.  The value of  isn’t really important to determining if the point is a maximum or a minimum so often we will not bother with finding a value for it.  On occasion we will need its value to help solve the system, but even in those cases we won’t use it past finding the point.


Example 2  Find the maximum and minimum of  subject to the constraint .



This one is going to be a little easier than the previous one since it only has two variables.  Also, note that it’s clear from the constraint that region of possible solutions lies on a disk of radius  which is a closed and bounded region and hence by the Extreme Value Theorem  we know that a minimum and maximum value must exist.


Here is the system that we need to solve.



Notice that, as with the last example, we can’t have  since that would not satisfy the first two equations.  So, since we know that  we can solve the first two equations for x and y respectively.  This gives,



Plugging these into the constraint gives,



We can solve this for .



Now, that we know  we can find the points that will be potential maximums and/or minimums.


If  we get,


and if  we get,



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.


Here are the minimum and maximum values of the function.



In the first two examples we’ve excluded  either for physical reasons or because it wouldn’t solve one or more of the equations.  Do not always expect this to happen.  Sometimes we will be able to automatically exclude a value of  and sometimes we won’t.


Let’s take a look at another example.


Example 3  Find the maximum and minimum values of  subject to the constraint .  Assume that .



First note that our constraint is a sum of three positive or zero number and it must be 1.  Therefore it is clear that our solution will fall in the range .  Therefore the solution must lie in a closed and bounded region and so by the Extreme Value Theorem  we know that a minimum and maximum value must exist.


Here is the system of equation that we need to solve.






Let’s start this solution process off by noticing that since the first three equations all have  they are all equal.  So, let’s start off by setting equations (10) and (11) equal.




So, we’ve got two possibilities here.  Let’s start off with by assuming that .  In this case we can see from either equation (10) or (11) that we must then have .  From equation (12) we see that this means that .  This in turn means that either  or


So, we’ve got two possible cases to deal with there.  In each case two of the variables must be zero.  Once we know this we can plug into the constraint, equation (13), to find the remaining value.




So, we’ve got two possible solutions  and .


Now let’s go back and take a look at the other possibility, .  We also have two possible cases to look at here as well.


This first case is .  In this case we can see from the constraint that we must have  and so we now have a third solution .


The second case is .  Let’s set equations (11) and (12) equal.



Now, we’ve already assumed that  and so the only possibility is that .  However, this also means that,



Using this in the constraint gives,


So, the next solution is .


We got four solutions by setting the first two equations equal. 


To completely finish this problem out we should probably set equations (10) and (12) equal as well as setting equations (11) and (12) equal to see what we get.  Doing this gives,



Both of these are very similar to the first situation that we looked at and we’ll leave it up to you to show that in each of these cases we arrive back at the four solutions that we already found.


So, we have four solutions that we need to check in the function to see whether we have minimums or maximums.



So, in this case the maximum occurs only once while the minimum occurs three times.


Note as well that we never really used the assumption that  in this problem.  This assumption is here mostly to make sure that we really do have a maximum and a minimum of the function.  Without this assumption it wouldn’t be too difficult to find points that give both larger and smaller values of the functions.  For example.




With these examples you can clearly see that it’s not too hard to find points that will give larger and smaller function values.  However, all of these examples required negative values of x, y and/or z to make sure we satisfy the constraint.  By eliminating these we will know that we’ve got minimum and maximum values by the Extreme Value Theorem.


To this point we’ve only looked at constraints that were equations.  We can also have constraints that are inequalities.  The process for these types of problems is nearly identical to what we’ve been doing in this section to this point.  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.


Let’s work an example to see how these kinds of problems work.


Example 4  Find the maximum and minimum values of  on the disk .



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.e. satisfy the constraint).  This is easy enough to do for this problem.  Here are the two first order partial derivatives.



So, the only critical point is  and it does satisfy the inequality.


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.


So, here is the system of equations that we need to solve.



From the first equation we get,



If we have  then the constraint gives us .


If we have  the second equation gives us,


The constraint then tells us that .


If we’d performed a similar analysis on the second equation we would arrive at the same points.


So, Lagrange Multipliers gives us four points to check : , , , and .


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.


We want to optimize  subject to the constraints  and .  The system that we need to solve in this case is,





So, in this case we get two Lagrange Multipliers.  Also, note that the first equation really is three equations as we saw in the previous examples.  Let’s see an example of this kind of optimization problem.


Example 5  Find the maximum and minimum of  subject to the constraints  and .



Verifying that we will have a minimum and maximum value here is a little trickier.  Clearly, because of the second constraint we’ve got to have .  With this in mind there must also be a set of limits on z in order to make sure that the first constraint is met.  If one really wanted to determine that range you could find the minimum and maximum values of  subject to  and you could then use this to determine the minimum and maximum values of z.  We won’t do that here.  The point is only to acknowledge that once again the possible solutions must lie in a closed and bounded region and so minimum and maximum values must exist by the Extreme Value Theorem.


Here is the system of equations that we need to solve.







First, let’s notice that from equation (16) we get .  Plugging this into equation (14) and equation (15) and solving for x and y respectively gives,



Now, plug these into equation (18).



So, we have two cases to look at here.  First, let’s see what we get when .  In this case we know that,


Plugging these into equation (17) gives,


So, we’ve got one solution.


Let’s now see what we get if we take .  Here we have,


Plugging these into equation (17) gives,


and there’s a second solution.


Now all that we need to is check the two solutions in the function to see which is the maximum and which is the minimum.




So, we have a maximum at  and a minimum at .

Absolute Minimums and Maximums Previous Section   Next Section Multiple Integrals (Introduction)
Partial Derivatives Previous Chapter   Next Chapter Multiple Integrals

Calculus III (Notes) / Applications of Partial Derivatives / Lagrange Multipliers    [Notes] [Practice Problems] [Assignment Problems]

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