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### Section 3-4 : Absolute Extrema

2. Find the absolute minimum and absolute maximum of $$f\left( {x,y} \right) = \left( {9{x^2} - 1} \right)\left( {1 + 4y} \right)$$ on the rectangle given by $$- 2 \le x \le 3$$, $$- 1 \le y \le 4$$.

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Start Solution

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

${f_x} = 18x\left( {1 + 4y} \right)\hspace{0.5in}{f_y} = 4\left( {9{x^2} - 1} \right)$ Show Step 2

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

\begin{align*} & {f_x} = 0\,\,\,\,:\,\,\,\,\,18x\left( {1 + 4y} \right) = 0\\ & {f_y} = 0\,\,\,\,:\,\,\,\,\,\,\,\,4\left( {9{x^2} - 1} \right) = 0\hspace{0.25in} \to \hspace{0.25in}\,\,\,x = \pm \frac{1}{3}\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.

$$\displaystyle x = \frac{1}{3}\,\,:\,\,\,6\left( {1 + 4y} \right) = 0\,\,\,\, \to \,\,\,\,\,y = - \frac{1}{4}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\frac{1}{3}, - \frac{1}{4}} \right)$$

$$\displaystyle x = - \frac{1}{3}\,\,:\,\,\, - 6\left( {1 + 4y} \right) = 0\,\,\,\, \to \,\,\,\,\,y = - \frac{1}{4}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( { - \frac{1}{3}, - \frac{1}{4}} \right)$$

Both of these critical points are in the region we are interested in and so we’ll need the function evaluated at both of them. Here are those values,

$f\left( {\frac{1}{3}, - \frac{1}{4}} \right) = 0\hspace{0.25in}\hspace{0.25in}f\left( { - \frac{1}{3}, - \frac{1}{4}} \right) = 0$ 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 rectangle can then be defined as follows.

Top :$$y = 4,\,\,\,\, - 2 \le x \le 3$$

Bottom : $$y = - 1,\,\,\,\, - 2 \le x \le 3$$

Right : $$x = 3,\,\,\,\,\,\, - 1 \le y \le 4$$

Left : $$x = - 2,\,\,\,\,\,\, - 1 \le y \le 4$$

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 : $$y = 4,\,\,\,\, - 2 \le x \le 3$$ .

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 $$- 2 \le x \le 3$$.

$g\left( x \right) = f\left( {x,4} \right) = 17\left( { - 1 + 9{x^2}} \right)$

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,

$g'\left( x \right) = 306x = 0\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}x = 0$

This critical point is in the interval we are working on so, this point as well as the $$x$$ limits for the top give the following three points that are potential absolute extrema for $$f\left( {x,y} \right)$$.

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

Recall that, in this step, we are assuming that $$y = 4$$! So, the next set of potential absolute extrema for $$f\left( {x,y} \right)$$ are then,

$f\left( {0,4} \right) = - 17\hspace{0.25in}\hspace{0.25in}f\left( { - 2,4} \right) = 595\hspace{0.25in}\,\,\,\,\,\,f\left( {3,4} \right) = 1360$ Show Step 5

Next, let’s check out the bottom : $$y = - 1,\,\,\,\, - 2 \le x \le 3$$. For this side we’ll need to identify possible absolute extrema of the following function on the interval $$- 2 \le x \le 3$$.

$g\left( x \right) = f\left( {x, - 1} \right) = - 3\left( { - 1 + 9{x^2}} \right)$

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

$g'\left( x \right) = - 54x = 0\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}x = 0$

This critical point is in the interval we are working on so, this point as well as the $$x$$ limits for the bottom give the following three points that are potential absolute extrema for $$f\left( {x,y} \right)$$.

$\left( {0, - 1} \right)\hspace{0.25in}\hspace{0.25in}\left( { - 2, - 1} \right)\hspace{0.25in}\,\,\,\,\,\,\,\left( {3, - 1} \right)$

Recall that, in this step, we are assuming that $$y = - 1$$! So, the next set of potential absolute extrema for $$f\left( {x,y} \right)$$ are then,

$f\left( {0, - 1} \right) = 3\hspace{0.5in}f\left( { - 2, - 1} \right) = - 105\hspace{0.5in}f\left( {3, - 1} \right) = - 240$ Show Step 6

Let’s now check out the right side : $$x = 3,\,\,\,\,\,\, - 1 \le y \le 4$$. For this side we’ll need to identify possible absolute extrema of the following function on the interval $$- 1 \le y \le 4$$.

$h\left( y \right) = f\left( {3,y} \right) = 80\left( {1 + 4y} \right)$

The derivative of the $$h\left( y \right)$$ from this step is,

$h'\left( y \right) = 320$

In this case there are no critical points of the function along this boundary. So, only the limits for the right side are potential absolute extrema for$$f\left( {x,y} \right)$$.

$\left( {3, - 1} \right)\hspace{0.25in}\hspace{0.25in}\left( {3,4} \right)$

Recall that, in this step, we are assuming that $$x = 3$$! Therefore, the next set of potential absolute extrema for $$f\left( {x,y} \right)$$ are then,

$f\left( {3, - 1} \right) = - 240\hspace{0.25in}\hspace{0.25in}f\left( {3,4} \right) = 1360$

Before proceeding to the next step let’s note that both of these points have already been listed in previous steps and so did not really need to be written down here. This will always happen with boundary points (as these are here). Boundary points will always show up in multiple boundary steps.

Show Step 7

Finally, let’s check out the left side : $$x = - 2,\,\,\,\,\,\, - 1 \le y \le 4$$. For this side we’ll need to identify possible absolute extrema of the following function on the interval $$- 1 \le y \le 4$$.

$h\left( y \right) = f\left( { - 2,y} \right) = 35\left( {1 + 4y} \right)$

The derivative of the $$h\left( y \right)$$ from this step is,

$h'\left( y \right) = 140$

In this case there are no critical points of the function along this boundary. So, we only the limits for the right side are potential absolute extrema for $$f\left( {x,y} \right)$$.

$\left( { - 2, - 1} \right)\hspace{0.25in}\hspace{0.25in}\left( { - 2,4} \right)$

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

$f\left( { - 2, - 1} \right) = - 105\hspace{0.25in}\hspace{0.25in}f\left( { - 2,4} \right) = 595$

As with the previous step both of these are boundary points and have appeared in previous steps. They were simply listed here for completeness.

Show Step 8

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( {0,4} \right) & = - 17\hspace{0.2in} & f\left( { - 2,4} \right) & = 595\hspace{0.2in} & f\left( {3,4} \right) & = 1360\\ f\left( {0, - 1} \right) & = 3\hspace{0.2in} & f\left( { - 2, - 1} \right) & = - 105\hspace{0.2in} & f\left( {3, - 1} \right) & = - 240\end{align*}

From this list we can see that the absolute maximum of the function will be 1360 which occurs at $$\left( {3,4} \right)$$ and the absolute minimum of the function will be -240 which occurs at $$\left( {3, - 1} \right)$$.