1.      A region in  is called closed if it includes its boundary.  A region is called open if it doesn’t include any of its boundary points.

2.      A region in  is called bounded if it can be completely contained in a disk.  In other words, a region will be bounded if it is finite.

If  is continuous in some closed, bounded set D in  then there are points in D,  and  so that  is the absolute maximum and  is the absolute minimum of the function in D.

1.      Find all the critical points of the function that lie in the region D and determine the function value at each of these points.

2.      Find all extrema of the function on the boundary.  This usually involves the Calculus I approach for this work.

3.      The largest and smallest values found in the first two steps are the absolute minimum and the absolute maximum of the function.

Example 1  Find the absolute minimum and absolute maximum of  on the rectangle given by  and .

Solution

Let’s first get a quick picture of the rectangle for reference purposes.

The boundary of this rectangle is given by the following conditions.

These will be important in the second step of our process.

We’ll start this off by finding all the critical points that lie inside the given rectangle.  To do this we’ll need the two first order derivatives.

Note that since we aren’t going to be classifying the critical points we don’t need the second order derivatives.  To find the critical points we will need to solve the system,

We can solve the second equation for y to get,

Plugging this into the first equation gives us,

This tells us that we must have  or .  Now, recall that we only want critical points in the region that we’re given.  That means that we only want critical points for which .  The only value of x that will satisfy this is the first one so we can ignore the last two for this problem.  Note however that a simple change to the boundary would include these two so don’t forget to always check if the critical points are in the region (or on the boundary since that can also happen).

Plugging  into the equation for y gives us,

The single critical point, in the region (and again, that’s important), is .  We now need to get the value of the function at the critical point.

Eventually we will compare this to values of the function found in the next step and take the largest and smallest as the absolute extrema of the function in the rectangle.

Now we have reached the long part of this problem.  We need to find the absolute extrema of the function along the boundary of the rectangle.  What this means is that we’re going to need to look at what the function is doing along each of the sides of the rectangle listed above.

Let’s first take a look at the right side.  As noted above the right side is defined by

Notice that along the right side we know that .  Let’s take advantage of this by defining a new function as follows,

Now, finding the absolute extrema of  along the right side will be equivalent to finding the absolute extrema of  in the range .  Hopefully you recall how to do this from Calculus I.  We find the critical points of  in the range  and then evaluate  at the critical points and the end points of the range of y’s.

Let’s do that for this problem.

 This is in the range and so we will need the following function evaluations.

Notice that, using the definition of  these are also function values for .

We can now do the left side of the rectangle which is defined by,

Again, we’ll define a new function as follows,

Notice however that, for this boundary, this is the same function as we looked at for the right side.  This will not always happen, but since it has let’s take advantage of the fact that we’ve already done the work for this function.  We know that the critical point is  and we know that the function value at the critical point and the end points are,

The only real difference here is that these will correspond to values of  at different points than for the right side.  In this case these will correspond to the following function values for .

We can now look at the upper side defined by,

We’ll again define a new function except this time it will be a function of x.

We need to find the absolute extrema of  on the range .  First find the critical point(s).

The value of this function at the critical point and the end points is,

and these in turn correspond to the following function values for  

Note that there are several “repeats” here.  The first two function values have already been computed when we looked at the right and left side.  This will often happen.

Finally, we need to take care of the lower side.  This side is defined by,

The new function we’ll define in this case is,

The critical point for this function is,

The function values at the critical point and the endpoint are,

and the corresponding values for  are,

The final step to this (long…) process is to collect up all the function values for  that we’ve computed in this problem.  Here they are,

The absolute minimum is at  since gives the smallest function value and the absolute maximum occurs at  and  since these two points give the largest function value.

Here is a sketch of the function on the rectangle for reference purposes.

Example 2  Find the absolute minimum and absolute maximum of   on the disk of radius 4,  

Solution

First note that a disk of radius 4 is given by the inequality in the problem statement.  The “less than” inequality is included to get the interior of the disk and the equal sign is included to get the boundary.  Of course, this also means that the boundary of the disk is a circle of radius 4.

Let’s first find the critical points of the function that lies inside the disk.  This will require the following two first order partial derivatives.

 To find the critical points we’ll need to solve the following system.

This is actually a fairly simple system to solve however.  The first equation tells us that  and the second tells us that .  So the only critical point for this function is  and this is inside the disk of radius 4.  The function value at this critical point is,

Now we need to look at the boundary.  This one will be somewhat different from the previous example.  In this case we don’t have fixed values of x and y on the boundary.  Instead we have,

We can solve this for  and plug this into the  in  to get a function of y as follows.

We will need to find the absolute extrema of this function on the range  (this is the range of y’s for the disk….).  We’ll first need the critical points of this function.

The value of this function at the critical point and the endpoints are,

Unlike the first example we will still need to find the values of x that correspond to these.  We can do this by plugging the value of y into our equation for the circle and solving for y.

The function values for  then correspond to the following function values for .

Note that the third one actually corresponded to two different values for  since that y also produced two different values of x.

So, comparing these values to the value of the function at the critical point of  that we found earlier we can see that the absolute minimum occurs at  while the absolute maximum occurs twice at  and .

Here is a sketch of the region for reference purposes.