Paul's Online Notes
Home / Calculus III / Applications of Partial Derivatives / Relative Minimums and Maximums
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 14.3 : Relative Minimums and Maximums

3. Find and classify all the critical points of the following function.

$f\left( {x,y} \right) = \left( {3x + 4{x^3}} \right)\left( {{y^2} + 2y} \right)$

Show All Steps Hide All Steps

Start Solution

We’re going to need a bunch of derivatives for this problem so let’s get those taken care of first.

Do not make these derivatives harder than really are! Do not multiply the function out! We just have a function of $$x$$’s times a function of $$y$$’s. Take advantage of that when doing the derivatives.

$\begin{array}{c}{f_x} = \left( {3 + 12{x^2}} \right)\left( {{y^2} + 2y} \right)\hspace{0.5in}{f_y} = \left( {3x + 4{x^3}} \right)\left( {2y + 2} \right)\\ {f_{x\,x}} = 24x\left( {{y^2} + 2y} \right)\hspace{0.25in}{f_{x\,y}} = \left( {3 + 12{x^2}} \right)\left( {2y + 2} \right)\hspace{0.25in}{f_{y\,y}} = 2\left( {3x + 4{x^3}} \right)\end{array}$ Show Step 2

Now, let’s find the critical points for this problem. That means solving the following system.

\begin{align*}{f_x} & = 0 : \,\,\left( {3 + 12{x^2}} \right)\left( {{y^2} + 2y} \right) = 0\\ {f_y} & = 0 : \left( {3x + 4{x^3}} \right)\left( {2y + 2} \right) = 0\end{align*}

We could start the solution process with either of these equations as both are pretty simple to solve. Let’s start with the first equation.

$\left( {3 + 12{x^2}} \right)\left( {{y^2} + 2y} \right) = \left( {3 + 12{x^2}} \right)\left( y \right)\left( {y + 2} \right) = 0\,\,\,\,\,\,\, \to \,\,\,\,\,y = 0,y = - 2,\,\,x = \pm \frac{1}{2}i$

Okay, we’ve got something to deal with at this point. We clearly get four different values to work with here. Two of them, however, are complex. One of the unspoken rules here is that we are only going to work with real values and so we will ignore any complex answers and work with only the real values.

So, we now have two possible values of $$y$$ so let’s plug each of them into the second equation as follows,

$$y = 0 : 2\left( {3x + 4{x^3}} \right) = 2x\left( {3 + 4{x^2}} \right) = 0\,\,\,\,\, \to \,\,\,\,\,x = 0,x = \pm \frac{{\sqrt 3 }}{2}i\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {0,0} \right)$$

$$y = - 2 : - 2\left( {3x + 4{x^3}} \right) = 2x\left( {3 + 4{x^2}} \right) = 0\,\,\,\,\, \to \,\,\,\,\,x = 0,x = \pm \frac{{\sqrt 3 }}{2}i\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {0, - 2} \right)$$

As with the first part of the solution process we only take the real values and so ignore the complex portions from this part as well.

In the previous two problems we made mention at this point to be careful and not just from up points for all possible combinations of the $$x$$ and $$y$$ values we have at this point.

One of the reasons that students often do that is because of problems like this one where it appears that we are doing just that. However, we haven’t just randomly formed all combinations here. It just so happened that when we assumed $$y = 0$$ and $$y = - 2$$ that we just happened to get the same value of $$x$$, $$x = 0$$. In general, this won’t happen and so do not read into this problem that we always just form all possible combinations of the $$x$$ and $$y$$ values to get the critical points for a function. We must always pay attention to the assumptions made at the start of each step.

So, in summary, this function has two critical points : $$\left( {0, - 2} \right),\,\,\left( {0,0} \right)$$.

Before proceeding with the next step we should note that there are multiple ways to solve this system. The process you used may not be the same as the one we used here. However, regardless of the process used to solve the system, the solutions should always be the same.

Show Step 3

Next, we’ll need the following,

\begin{align*}D\left( {x,y} \right) & = {f_{x\,x}}{f_{y\,y}} - {\left[ {{f_{x\,y}}} \right]^2}\\ & = \left[ {24x\left( {{y^2} + 2y} \right)} \right]\left[ {2\left( {3x + 4{x^3}} \right)} \right] - {\left[ {\left( {3 + 12{x^2}} \right)\left( {2y + 2} \right)} \right]^2}\\ & = 48x\left( {3x + 4{x^3}} \right)\left( {{y^2} + 2y} \right) - {\left( {3 + 12{x^2}} \right)^2}{\left( {2y + 2} \right)^2}\end{align*} Show Step 4

With $$D\left( {x,y} \right)$$ we can now classify each of the critical points as follows.

\begin{align*} & \left( {0, - 2} \right) & : \hspace{0.25in} & D\left( {0, - 2} \right) = - 36 < 0 & \hspace{0.25in}& {\mbox{Saddle Point}}\\ & \left( {0,0} \right) & : \hspace{0.25in} & D\left( {0,0} \right) = - 36 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\end{align*}

Don’t always expect every problem to have at least one relative extrema. As this example has shown it is completely possible to have only saddle points.