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In that example you can see some of the ideas you might need to do in order to find the optimal value.So, let’s get the derivative and find the critical points.Before we give a summary of this method let’s discuss the continuity requirement a little. If you can do one you can do the other as well. In Example 3, on the other hand, we were trying to optimize the volume and the surface area was the constraint.So, we only have a single critical point to deal with here and notice that 6.2035 is the only value for which the derivative will be zero and hence the only place (with \(r > 0\) of course) that the derivative may change sign. It is however easy to confuse the two if you just skim the problem so make sure you carefully read the problem first!In this problem the constraint is the volume and we want to minimize the amount of material used.
The volume is just the area of each of the disks times the height. Here are those function evaluations.So, if we take \(h = 1.9183\) we get a maximum volume.Solving the constraint for \(h\) and plugging into the equation for the printed area gives, What it does do is allow us to potentially exclude values and knowing this can simplify our work somewhat and so is not a bad thing to do.Similarly, if we know that to the left of \(x = c\) the function is always decreasing and to the right of \(x = c\) the function is always increasing then the absolute minimum of \(f\left( x \right)\) in \(I\) will occur at \(x = c\).In the previous problem we used the method from the Finding Absolute Extrema section to find the maximum value of the function we wanted to optimize. In some cases, the method we use will be the only method we could use, in others it will be the easiest method to use and in others it will simply be the method we chose to use for that example. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. Do not however get into the habit of just excluding any negative critical point. One possibility is to change the value at one site at a time, while leaving all other pixels unchanged. There are a couple of examples in the next two sections with more than one critical point including one in the next section mentioned above in which none of the methods discussed above easily work. Recall that in order to use this method the interval of possible values of the independent variable in the function we are optimizing, let’s call it \(I\), must have finite endpoints. The only problem is it don't show a new equation number for each line. The definition of “small” perturbation in step 2 depends on the particular optimization problem [32].

However, we will always need to use some method for making sure that our answer is in fact that optimal value that we’re after.Note that if you think of a cylinder of height \(h\) and radius \(r\) as just a bunch of disks/circles of radius \(r\) stacked on top of each other the equations for the surface area and volume are pretty simple to remember. Since we are after the absolute maximum we know that a maximum (of any kind) can’t occur at relative minimums and so we immediately know that we can exclude these points from further consideration.
It is considered a basic management technique that can be viewed as a loop of measurement, improvement and measurement. The smallest \(h\) can be is \(h = 0\) even though this doesn’t make much sense as we won’t get a box in this case. We don’t have a cost here, but if you think about it the cost is nothing more than the amount of material used times a cost and so the amount of material and cost are pretty much tied together.

First, notice that \(w = 0\) is not a critical point. The constraint is simply the size of the piece of cardboard and has already been factored into the figure above.