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### Section 3-5 : Lagrange Multipliers

3. Find the maximum and minimum values of $$f\left( {x,y,z} \right) = {y^2} - 10z$$ subject to the constraint $${x^2} + {y^2} + {z^2} = 36$$.

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Start Solution

Before proceeding with the problem let’s note because our constraint is the sum of three terms that are squared (and hence positive) the largest possible range of $$x$$ is $$- 6 \le x \le 6$$ (the largest values would occur if $$y = 0$$ and $$z = 0$$). Likewise, we’d get the same ranges for both $$y$$ and $$z$$.

Note that, at this point, we don’t know if $$x$$, $$y$$ or $$z$$ will actually be the largest possible value. At this point we are simply acknowledging what they are. What this allows us to say is that whatever our answers will be they must occur in these bounded ranges and hence by the Extreme Value Theorem we know that absolute extrema will occur for this problem.

This step is an important (and often overlooked) step in these problems. It always helps to know that absolute extrema exist prior to actually trying to find them!

Show Step 2

The first actual step in the solution process is then to write down the system of equations we’ll need to solve for this problem.

\begin{align*}0 & = 2x\lambda \\ 2y & = 2y\lambda \\ - 10 & = 2z\lambda \\ {x^2} + {y^2} + {z^2} &= 36\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.

For this system let’s start with the third equation and note that because the left side is -10, or more importantly can never by zero, we can see that we must therefore have $$z \ne 0$$ and $$\lambda \ne 0$$. The fact that $$\lambda$$ can’t be zero is really important for this problem.

Show Step 4

Okay, because we now know that $$\lambda \ne 0$$ we can see that the only way for the first equation to be true is to have $$x = 0$$.

Therefore, no matter what else is going on with $$y$$ and $$z$$ in this problem we must always have $$x = 0$$ and we’ll need to keep that in mind.

Show Step 5

Next, let’s take a look at the second equation. A quick rewrite of this equation gives,

$2y - 2y\lambda = 2y\left( {1 - \lambda } \right) = 0\hspace{0.25in} \to \hspace{0.25in}y = 0\,\,\,\,{\mbox{or}}\,\,\,\,\lambda = 1$ Show Step 6

We now have two possibilities from Step 4. Either $$y = 0$$ or $$\lambda = 1$$. We’ll need to go through both of these possibilities and see what we get.

Let’s start by assuming that $$y = 0$$ and recall from Step 3 that we also know that $$x = 0$$. In this case we can plug these values into the constraint to get,

${z^2} = 36\hspace{0.25in} \to \hspace{0.25in}z = \pm 6$

Therefore, from this part we get two points that are potential absolute extrema,

$\left( {0,0, - 6} \right)\hspace{0.5in}\left( {0,0,6} \right)$ Show Step 7

Next, let’s assume that $$\lambda = 1$$. If we head back to the third equation we can see that we now have,

$- 10 = 2z\hspace{0.25in} \to \hspace{0.25in}z = - 5$

So, under this assumption we must have $$z = - 5$$ and recalling once more from Step 3 that we have $$x = 0$$ we can now plug these into the constraint to get,

${y^2} + 25 = 36\hspace{0.25in} \to \hspace{0.25in}{y^2} = 11\hspace{0.25in} \to \hspace{0.25in}y = \pm \sqrt {11}$

So, this part gives us two more points that are potential absolute extrema,

$\left( {0, - \sqrt {11} , - 5} \right)\hspace{0.5in}\left( {0,\sqrt {11} , - 5} \right)$ Show Step 8

In total, it looks like we have four 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. Here are those function evaluations.

$f\left( {0, - \sqrt {11} , - 5} \right) = 61\,\,\,\,\,\,\,\,\,f\left( {0,\sqrt {11} , - 5} \right) = 61\,\,\,\,\,\,\,\,\,\,f\left( {0,0, - 6} \right) = 60\,\,\,\,\,\,\,\,\,f\left( {0,0,6} \right) = - 60$

The absolute maximum is then 61 which occurs at $$\left( {0, - \sqrt {11} , - 5} \right)$$ and $$\left( {0,\sqrt {11} , - 5} \right)$$. The absolute minimum is -60 which occurs at $$\left( {0,0,6} \right)$$. Do not get excited about the absolute extrema occurring at multiple points. That will happen on occasion with these problems.