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### Section 3-5 : Lagrange Multipliers

4. Find the maximum and minimum values of $$f\left( {x,y,z} \right) = xyz$$ subject to the constraint $$x + 9{y^2} + {z^2} = 4$$. Assume that $$x \ge 0$$ for this problem. Why is this assumption needed?

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Start Solution

Before proceeding with the solution to this problem let’s address why the assumption that $$x \ge 0$$ is needed for this problem.

The answer is simple. Without that assumption this function will not have absolute extrema.

If there are no restrictions on $$x$$ then we could make $$x$$ as large and negative as we wanted to and we could still meet the constraint simply by chose a very large $$y$$ and/or $$z$$. Note as well that because $$y$$ and $$z$$ are both squared we could chose them to be either negative or positive.

If we took our choices for $$x$$, $$y$$ and $$z$$ and plugged them into the function then the function would be similarly large. Also, the larger we chose $$x$$ the larger we’d need to choose appropriate $$y$$ and/or $$z$$ and hence the larger our function would become. Finally, as noted above because we could chose $$y$$ and $$z$$ to be either positive or negative we could force the function to be either positive or negative with appropriate choices of signs for $$y$$ and $$z$$.

In other words, if we have no restriction on $$x$$, we can make the function arbitrarily large in a positive and negative sense and so this function would not have absolute extrema.

On the other hand, if we put on the restriction on $$x$$ that we have we now have the sum of three positive terms that must equal four. This in turn leads to the following largest possible values of the three variables in the problem.

$0 \le x \le 4\hspace{0.5in} - \frac{2}{3} \le y \le \frac{2}{3}\hspace{0.5in} - 2 \le z \le 2$

The largest value of $$x$$ and the extreme values of $$y$$ and $$z$$ would occur when the other two variables are zero and in general there is no way to know ahead of time if any of the variables will in fact take on their largest possible values. However, what we can say now is that because all of our variables are bounded then by the Extreme Value Theorem we know that absolute extrema will occur for this problem.

Note as well that all we really need here is a lower limit for $$x$$. It doesn’t have to be zero that just makes the above analysis a little bit easier. We could have used the restriction that $$x \ge - 8$$ if we’d wanted to. With this restriction we’d still have a bounded set of ranges for $$x$$, $$y$$ and $$z$$ and so the function would still have absolute extrema.

This problem shows just why this step is so important for these problems. If this problem did not have a restriction on $$x$$ and we neglected to do this step we’d get the (very) wrong answer! We could still go through the process below and we’d get values that would appear to be absolute extrema. However, as we’ve shown above without any restriction on $$x$$ the function would not have absolute extrema.

The issue here is that the Lagrange multiplier process itself is not set up to detect if absolute extrema exist or not. Before we even start the process we need to first make sure that the values we get out of the process will in fact be absolute extrema (i.e. we need to verify that absolute extrema exist).

Show Step 2

The first step here is to write down the system of equations we’ll need to solve for this problem.

\begin{align*}yz & = \lambda \\ xz & = 18y\lambda \\ xy & = 2z\lambda \\ x + 9{y^2} + {z^2} & = 4\end{align*} Show Step 3

For most of these systems there are a multitude of solution methods that we can use to find a solution. Some may be harder than other, but unfortunately, there will often be no way of knowing which will be “easy” and which will be “hard” until you start the solution process.

Do not be afraid of these systems. They are probably unlike anything you’ve ever really been asked to solve up to this point. Most of the systems can be solved using techniques that you already know and aren’t really as “bad” as they may appear at first glance. Some do require some additional techniques and can be quite messy but for the most part still involve techniques that you do know how to use, you just may not have ever seen them done in the context of solving systems of equations.

With this system let’s start out by multiplying the first equation by $$x$$, multiplying the second equation by $$y$$ and multiplying the third equation by $$z$$. Doing this gives the following “new” system of equations.

\begin{align*}xyz & = x\lambda \\ xyz & = 18{y^2}\lambda \\ xyz & = 2{z^2}\lambda \\ x + 9{y^2} + {z^2} & = 4\end{align*}

Let’s also note that the constraint won’t be true if all three variables are zero simultaneously. One or two of the variables can be zero but we can’t have all three be zero.

Show Step 4

Now, let’s set the first and second equations from Step 3 equal. Doing this gives,

$x\lambda = 18{y^2}\lambda \hspace{0.25in} \to \hspace{0.25in}\left( {x - 18{y^2}} \right)\lambda = 0\hspace{0.25in} \to \hspace{0.25in}x = 18{y^2}\,\,\,{\mbox{or}}\,\,\,\lambda = 0$

Let’s also set the second and third equation from Step 3 equal. Doing this gives,

$18{y^2}\lambda = 2{z^2}\lambda \hspace{0.25in} \to \hspace{0.25in}\left( {18{y^2} - 2{z^2}} \right)\lambda = 0\hspace{0.25in}\, \to \hspace{0.25in}{z^2} = 9{y^2}\,\,\,{\mbox{or}}\,\,\,\,\lambda = 0$ Show Step 5

Okay, from Step 4 we have two possibilities. Either $$\lambda = 0$$ or we have $$x = 18{y^2}$$ and $${z^2} = 9{y^2}$$.

Let’s take care of the first possibility, $$\lambda = 0$$. If we go back to the original system this assumption gives us the following system.

\begin{align*}yz & = 0\hspace{0.25in} \to \hspace{0.25in}y = 0\,\,\,\,{\mbox{or}}\,\,\,\,z = 0\\ xz & = 0\hspace{0.25in} \to \hspace{0.25in}x = 0\,\,\,\,{\mbox{or}}\,\,\,\,z = 0\\ xy & = 0\hspace{0.25in} \to \hspace{0.25in}x = 0\,\,\,\,{\mbox{or}}\,\,\,\,y = 0\\ x + 9{y^2} + {z^2} & = 4\end{align*} Show Step 6

We have all sorts of possibilities from Step 5. From the first equation we have two possibilities. Let’s start with $$y = 0$$. Since the third equation from Step 5 won’t really tell us anything (after all it is now 0 = 0) let’s move to the second equation. In this case we get either $$x = 0$$ or $$z = 0$$.

Recall that at the end of the third step we noticed that we can’t have all three of the variables be zero but we could have two of them be zero. So, this leads to the following two cases that we can plug into the constraint to find the value of the third variable.

\begin{align*}y = 0,x = 0\,\,\,\,\, & :\,\,\,\,\,\,\,{z^2} = 4\,\,\,\,\,\,\, \to \hspace{0.25in}\,z = \pm 2\\ y = 0,z = 0\,\,\,\,\, & :\,\,\,\,\,\,\,x = 4\end{align*}

So, this gives us the following three potential absolute extrema.

$\left( {0,0, - 2} \right)\hspace{0.25in}\,\left( {0,0,2} \right)\hspace{0.25in}\,\left( {4,0,0} \right)$

Next, let’s take a look at the second possibility from the first equation in Step 5, $$z = 0$$. In this case the second equation will be 0 = 0 and so will not be of any use. The third however, has the possibilities of $$x = 0$$ or $$y = 0$$. The second of these was already addressed above so all we need to look at is,

$z = 0,x = 0\,\,\,\,\,:\,\,\,\,\,\,\,9{y^2} = 4\,\,\,\,\,\,\, \to \hspace{0.25in}\,y = \pm \frac{2}{3}$

This leads to two more potential absolute extrema.

$\left( {0, - \frac{2}{3},0} \right)\hspace{0.25in}\,\left( {0,\frac{2}{3},0} \right)$ We could now go back and start with the second or third equation but if we did that you’d just end up with the above possibilities (you might want to verify that for yourself…). Therefore, we get a total of five possible absolute extrema from this Step. They are,

$\left( {0,0, \pm 2} \right)\hspace{0.25in}\,\left( {0, \pm \frac{2}{3},0} \right)\hspace{0.25in}\,\left( {4,0,0} \right)$

We made heavy use of the “$$\pm$$” notation here to simplify things a little bit. It’s not required but will make the rest of the work with these points a little easier as we’ll see eventually.

Show Step 7

Now, way back in Step 5 we had another possibility : $$x = 18{y^2}$$ and $${z^2} = 9{y^2}$$. We have to now take a look at this case. In this case we can plug each of these directly into the constraint to get the following,

$18{y^2} + 9{y^2} + 9{y^2} = 36{y^2} = 4\hspace{0.25in}\,\, \to \hspace{0.25in}\,\,y = \pm \frac{1}{3}$

Now we can go back to the two assumptions we started this step off with to get,

$x = 18\left( {\frac{1}{9}} \right) = 2\hspace{0.5in}{z^2} = 9\left( {\frac{1}{9}} \right) = 1\hspace{0.25in}\,\, \to \hspace{0.25in}\,\,z = \pm 1$

Now, in most cases, we can’t just “mix and match” all the values of $$x$$, $$y$$ and $$z$$ to from points. In this case however, we can do exactly that. The $$x = 2$$ will arise regardless of the sign on $$y$$ because of the $${y^2}$$ in the $$x$$ assumption. Likewise, because of the $${y^2}$$ in the $$z$$ assumption each of the $$z$$’s can arise for either $$y$$ and so we get all combinations of $$x$$, $$y$$ and $$z$$ for points in this case.

Therefore, we get the following four possible absolute extrema from this step.

$\left( {2, - \frac{1}{3}, - 1} \right)\hspace{0.25in}\,\,\,\left( {2, - \frac{1}{3},1} \right)\hspace{0.25in}\,\,\,\left( {2,\frac{1}{3}, - 1} \right)\hspace{0.25in}\,\,\,\left( {2,\frac{1}{3},1} \right)$ Show Step 8

In total, it looks like we have nine points that can potentially be absolute extrema. So, to determine the absolute extrema all we need to do is evaluate the function at each of these points and with nine points that seems like a lot of work.

However, in this case, it’s actually quite simple. Recall that the function we’re evaluating is $$f\left( {x,y,z} \right) = xyz$$. First, this means that if even one of the variables is zero the whole function will be zero. Therefore, the function evaluations for the five points from Step 6 all give,

$f\left( {0,0, \pm 2} \right) = f\left( {0, \pm \frac{2}{3},0} \right) = f\,\left( {4,0,0} \right) = 0$

Note the usage of the “$$\pm$$” notation to “simplify” the work here as well.

Now, the potential points from Step 7 are all the same values, with the exception of signs changing occasionally on the $$y$$ and $$z$$. That means that the function value here will be either $$- \frac{2}{3}$$ or $$\frac{2}{3}$$ depending on the number of minus signs in the point. So again, not a lot of effort to compute these function values. Here are the evaluations for the points from Step 7.

$f\left( {2, - \frac{1}{3},1} \right) = f\left( {2,\frac{1}{3}, - 1} \right) = - \frac{2}{3}\hspace{0.5in}f\left( {2, - \frac{1}{3}, - 1} \right) = f\left( {2,\frac{1}{3},1} \right) = \frac{2}{3}$

The absolute maximum is then $$\frac{2}{3}$$ which occurs at $$\left( {2, - \frac{1}{3}, - 1} \right)$$ and $$\left( {2,\frac{1}{3},1} \right)$$. The absolute minimum is $$- \frac{2}{3}$$ which occurs at $$\left( {2, - \frac{1}{3},1} \right)$$ and $$\left( {2,\frac{1}{3}, - 1} \right)$$. Do not get excited about the absolute extrema occurring at multiple points. That will happen on occasion with these problems.

Before leaving this problem we should note that some of the solution processes for the systems that arise with Lagrange multipliers can be quite involved. It can be easy to get lost in the details of the solution process and forget to go back and take care of one or more possibilities. You need to always be very careful and before finishing a problem go back and make sure that you’ve dealt with all the possible solution paths in the problem.