Paul's Online Notes
Home / Calculus III / Applications of Partial Derivatives / Absolute 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 3-4 : Absolute Extrema

1. Find the absolute minimum and absolute maximum of $$f\left( {x,y} \right) = 192{x^3} + {y^2} - 4x{y^2}$$ on the triangle with vertices $$\left( {0,0} \right)$$, $$\left( {4,2} \right)$$ and $$\left( { - 2,2} \right)$$.

Show All Steps Hide All Steps

Start Solution

We’ll need the first order derivatives to start the problem off. Here they are,

${f_x} = 576{x^2} - 4{y^2}\hspace{0.5in}{f_y} = 2y - 8xy$ Show Step 2

We need to find the critical points for this problem. That means solving the following system.

\begin{align*} & {f_x} = 0\,\,\,\,:\,\,\,\,\,576{x^2} - 4{y^2} = 0\\ & {f_y} = 0\,\,\,\,:\,\,\,\,\,\,\,\,2y\left( {1 - 4x} \right) = 0\hspace{0.25in} \to \hspace{0.25in}\,\,\,y = 0\,\,\,{\mbox{or}}\,\,\,x = \frac{1}{4}\end{align*}

So, we have two possible options from the second equation. We can plug each into the first equation to get the critical points for the equation.

$$y = 0\,\,:\,\,\,576{x^2} = 0\,\,\,\, \to \,\,\,\,\,x = 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {0,0} \right)$$

$$x = \frac{1}{4}\,\,:\,\,\,\,36 - 4{y^2} = 0\,\,\,\,\, \to \,\,\,\,\,y = \pm 3\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\frac{1}{4},3} \right)\,\,\,\,\,{\mbox{and}}\,\,\,\,\,\,\left( {\frac{1}{4}, - 3} \right)$$

Okay, we have the three critical points listed above. Also recall that we only use critical points that are actually in the region we are working with. In this case, the last two have $$y$$ values that clearly are out of the region (we’ve sketched the region in the next step if you aren’t sure you believe this!) and so we can ignore them.

Therefore, the only critical point from this list that we need to use is the first. Note as well that, in this case, this also happens to be one of the points that define the boundary of the region. This will happen on occasion but won’t always.

So, we’ll need the function value for the only critical point that is actually in our region. Here is that value,

$f\left( {0,0} \right) = 0$ Show Step 3

Now, we know that absolute extrema can occur on the boundary. So, let’s start off with a quick sketch of the region we’re working on.

Each of the sides of the triangle can then be defined as follows.

Top : $$y = 2,\,\,\, - 2 \le x \le 4$$

Right : $$y = \frac{1}{2}x,\,\,\,0 \le x \le 4$$

Left : $$y = - x,\,\,\, - 2 \le x \le 0$$

Now we need to analyze each of these sides to get potential absolute extrema for $$f\left( {x,y} \right)$$ that might occur on the boundary.

Show Step 4

Let’s first check out the top :$$y = 2,\,\,\, - 2 \le x \le 4$$.

We’ll need to identify the points along the top that could be potential absolute extrema for $$f\left( {x,y} \right)$$. This, in essence, requires us to find the potential absolute extrema of the following equation on the interval $$- 2 \le x \le 4$$.

$g\left( x \right) = f\left( {x,2} \right) = 192{x^3} - 16x + 4$

This is really nothing more than a Calculus I absolute extrema problem so we’ll be doing the work here without a lot of explanation. If you don’t recall how to do these kinds of problems you should read through that section in the Calculus I material.

The critical point(s) for $$g\left( x \right)$$ are,

$g'\left( x \right) = 576{x^2} - 16 = 0\hspace{0.5in}\to \hspace{0.5in}x = \pm \frac{1}{6}$

So, these two points as well as the $$x$$ limits for the top give the following four points that are potential absolute extrema for $$f\left( {x,y} \right)$$.

$\left( {\frac{1}{6},2} \right)\hspace{0.5in}\left( { - \frac{1}{6},2} \right)\hspace{0.5in}\left( { - 2,2} \right)\hspace{0.5in}\left( {4,2} \right)$

Recall that, in this step, we are assuming that $$y = 2$$! So, the next set of potential absolute extrema for $$f\left( {x,y} \right)$$ are then,

$f\left( {\frac{1}{6},2} \right) = \frac{{20}}{9}\hspace{0.5in}f\left( { - \frac{1}{6},2} \right) = \frac{{52}}{9}\hspace{0.5in}f\left( { - 2,2} \right) = - 1,500\hspace{0.5in}f\left( {4,2} \right) = 12,228$ Show Step 5

Next let’s check out the right side :$$y = \frac{1}{2}x,\,\,\,0 \le x \le 4$$. For this side we’ll need to identify possible absolute extrema of the following function on the interval $$0 \le x \le 4$$.

$g\left( x \right) = f\left( {x,\frac{1}{2}x} \right) = \frac{1}{4}{x^2} + 191{x^3}$

The critical point(s) for the $$g\left( x \right)$$ from this step are,

$g'\left( x \right) = \frac{1}{2}x + 573{x^2} = x\left( {\frac{1}{2} + 573x} \right) = 0\hspace{0.5in}\to \hspace{0.5in}x = 0,\,\,\,\,x = - \frac{1}{{1146}}$

Now, recall what we are restricted to the interval $$0 \le x \le 4$$ for this portion of the problem and so the second critical point above will not be used as it lies outside this interval.

So, the single point from above that is in the interval $$0 \le x \le 4$$ as well as the $$x$$ limits for the right give the following two points that are potential absolute extrema for $$f\left( {x,y} \right)$$.

$\left( {0,0} \right)\hspace{0.5in}\left( {4,2} \right)$

Recall that, in this step, we are assuming that $$y = \frac{1}{2}x$$! Also note that, in this case, one of the critical points ended up also being one of the endpoints.

Therefore, the next set of potential absolute extrema for $$f\left( {x,y} \right)$$ are then,

$f\left( {0,0} \right) = 0\hspace{0.5in}f\left( {4,2} \right) = 12,228$

Before proceeding to the next step note that both of these have already appeared in previous steps. This will happen on occasion but we can’t, in many cases, expect this to happen so we do need to go through and do the work for each boundary.

The main exception to this is usually the endpoints of our intervals as they will always be shared in two of the boundary checks and so, once done, don’t really need to be checked again. We just included the endpoints here for completeness.

Show Step 6

Finally, let’s check out the left side :$$y = - x,\,\,\, - 2 \le x \le 0$$. For this side we’ll need to identify possible absolute extrema of the following function on the interval $$- 2 \le x \le 0$$.

$g\left( x \right) = f\left( {x, - x} \right) = {x^2} + 188{x^3}$

The critical point(s) for the $$g\left( x \right)$$ from this step are,

$g'\left( x \right) = 2x + 564{x^2} = 2x\left( {1 + 282x} \right) = 0\hspace{0.5in}\to \hspace{0.5in}x = 0,\,\,\,\,x = - \frac{1}{{282}}$

Both of these are in the interval $$- 2 \le x \le 0$$ that we are restricted to for this portion of the problem.

So, the two points from above as well as the $$x$$ limits for the right give the following three points that are potential absolute extrema for $$f\left( {x,y} \right)$$.

$\left( { - \frac{1}{{282}},\frac{1}{{282}}} \right)\hspace{0.5in}\left( {0,0} \right)\hspace{0.5in}\left( { - 2,2} \right)$

Recall that, in this step we are assuming that $$y = - x$$! Also note that, in this case, one of the critical points ended up also being one of the endpoints.

Therefore, the next set of potential absolute extrema for $$f\left( {x,y} \right)$$ are then,

$f\left( { - \frac{1}{{282}},\frac{1}{{282}}} \right) = \frac{1}{{238,572}}\hspace{0.5in}f\left( {0,0} \right) = 0\hspace{0.5in}f\left( { - 2,2} \right) = - 1,500$

As with the previous step we can note that both of the end points above have already occurred previously in the problem and didn’t really need to be checked here. They were just included for completeness.

Show Step 7

Okay, in summary, here are all the potential absolute extrema and their function values for this function on the region we are working on.

\begin{align*}f\left( {\frac{1}{6},2} \right) & = \frac{{20}}{9}\hspace{0.2in} & f\left( { - \frac{1}{6},2} \right) & = \frac{{52}}{9} & \hspace{0.2in}f\left( { - 2,2} \right) = - 1,500\hspace{0.5in}f\left( {4,2} \right) = 12,228\\ f\left( {0,0} \right) & = 0 \hspace{0.2in} & f\left( { - \frac{1}{{282}},\frac{1}{{282}}} \right) & = \frac{1}{{238,572}} & \end{align*}

From this list we can see that the absolute maximum of the function will be 12,228 which occurs at $$\left( {4,2} \right)$$ and the absolute minimum of the function will be -1,500 which occurs at $$\left( { - 2,2} \right)$$.