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In this section we are going to look at optimization
problems. In optimization problems we
are looking for the largest value or the smallest value that a function can
take. We saw how to one kind of
optimization problem in the Absolute Extrema
section where we found the largest and smallest value that a function would
take on an interval.
In this section we are going to look at another type of
optimization problem. Here we will be
looking for the largest or smallest value of a function subject to some kind of
constraint. The constraint will be some
condition (that can usually be described by some equation) that must
absolutely, positively be true no matter what our solution is. On occasion, the constraint will not be
easily described by an equation, but in these problems it will be easy to deal
with as we’ll see.
This section is generally one of the more difficult for
students taking a Calculus course. One
of the main reasons for this is that a subtle change of wording can completely
change the problem. There is also the
problem of identifying the quantity that we’ll be optimizing and the quantity
that is the constraint and writing down equations for each.
The first step in all of these problems should be to very
carefully read the problem. Once you’ve
done that the next step is to identify the quantity to be optimized and the
constraint.
In identifying the constraint remember that the constraint
is something that must true regardless of the solution. In almost every one of the problems we’ll be
looking at here one quantity will be clearly indicated as having a fixed value
and so must be the constraint. Once
you’ve got that identified the quantity to be optimized should be fairly simple
to get. It is however easy to confuse
the two if you just skim the problem so make sure you carefully read the
problem first!
Let’s start the section off with a simple problem to
illustrate the kinds of issues will be dealing with here.
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Example 1 We
need to enclose a field with a fence.
We have 500 feet of fencing material and a building is on one side of
the field and so won’t need any fencing.
Determine the dimensions of the field that will enclose the largest
area.
Solution
In all of these problems we will have two functions. The first is the function that we are
actually trying to optimize and the second will be the constraint. Sketching the situation will often help us
to arrive at these equations so let’s do that.
In this problem we want to maximize the area of a field
and we know that will use 500 ft of fencing material. So, the area will be the function we are
trying to optimize and the amount of fencing is the constraint. The two
equations for these are,
Okay, we know how to find the largest or smallest value of
a function provided it’s only got a single variable. The area function (as well as the
constraint) has two variables in it and so what we know about finding
absolute extrema won’t work. However,
if we solve the constraint for one of the two variables we can substitute
this into the area and we will then have a function of a single variable.
So, let’s solve the constraint for x. Note that we could have
just as easily solved for y but
that would have led to fractions and so, in this case, solving for x will probably be best.
Substituting this into the area function gives a function
of y.
Now we want to find the largest value this will have on
the interval [0,250]. Note that the
interval corresponds to taking (i.e.
no sides to the fence) and (i.e.
only two sides and no width, also if there are two sides each must be 250 ft
to use the whole 500ft…).
Note that the endpoints of the interval won’t make any
sense from a physical standpoint if we actually want to enclose some area
because they would both give zero area.
They do, however, give us a set of limits on y and so the Extreme Value Theorem
tells us that we will have a maximum value of the area somewhere between the
two endpoints. Having these limits
will also mean that we can use the process we discussed in the Finding Absolute
Extrema section earlier in the chapter to find the maximum value of the
area.
So, recall that the maximum value of a continuous function
(which we’ve got here) on a closed interval (which we also have here) will
occur at critical points and/or end points.
As we’ve already pointed out the end points in this case will give
zero area and so don’t make any sense.
That means our only option will be the critical points.
So let’s get the derivative and find the critical points.
Setting this equal to zero and solving gives a lone
critical point of . Plugging this into the area gives an area
of 31250 ft2. So according
to the method from Absolute Extrema section this must be the largest possible
area, since the area at either endpoint is zero.
Finally, let’s not forget to get the value of x and then we’ll have the dimensions
since this is what the problem statement asked for. We can get the x by plugging in our y
into the constraint.
The dimensions of the field that will give the largest
area, subject to the fact that we used exactly 500 ft of fencing material,
are 250 x 125.
Don’t forget to actually read the problem and give the
answer that was asked for. These types
of problems can take a fair amount of time/effort to solve and it’s not hard
to sometimes forget what the problem was actually asking for.
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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. However, as we’ll see in later
examples we won’t always have easy to find endpoints and/or dealing with the
endpoints may not be easy to deal with.
Not only that, but this method requires that the function we’re
optimizing be continuous on the interval we’re looking at, including the
endpoints, and that may not always be the case.
So, before proceeding with the anymore examples let’s spend
a little time discussing some methods for determining if our solution is in
fact the absolute minimum/maximum value that we’re looking for. In some examples all of these will work while
in others one or more won’t be all that useful.
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.
Method 1 : Use the method used in Finding Absolute Extrema.
This is the method used in the first example above. Recall that in order to use this method the
range of possible optimal values, let’s call it I, must have finite endpoints.
Also, the function we’re optimizing (once it’s down to a single
variable) must be continuous on I,
including the endpoints. If these
conditions are met then we know that the optimal value, either the maximum or
minimum depending on the problem, will occur at either the endpoints of the
range or at a critical point that is inside the range of possible solutions.
There are two main issues that will often prevent this
method from being used however. First,
not every problem will actually have a range of possible solutions that have
finite endpoints at both ends. We’ll see
at least one example of this as we work through the remaining examples. Also, many of the functions we’ll be
optimizing will not be continuous once we reduce them down to a single variable
and this will prevent us from using this method.
Method 2 :
Use a variant of the First Derivative Test.
In this method we also will need a range of possible optimal
values, I. However, in this case, unlike the previous
method the endpoints do not need to be finite.
Also, we will need to require that the function be continuous on the
interior I and we will only need the
function to be continuous at the end points if the endpoint is finite and the
function actually exists at the endpoint.
We’ll see several problems where the function we’re optimizing doesn’t
actually exist at one of the endpoints.
This will not prevent this method from being used.
Let’s suppose that is a critical point of the function we’re
trying to optimize, . We already know from the First Derivative
Test that if immediately to the left of (i.e.
the function is increasing immediately to the left) and if immediately to the right of (i.e.
the function is decreasing immediately to the right) then will be a relative maximum for .
Now, this does not mean that the absolute maximum of will occur at . However, suppose that we knew a little bit
more information. Suppose that in fact
we knew that for all x
in I such that . Likewise, suppose that we knew that for all x
in I such that . In this case we know that to the left of ,
provided we stay in I of course, the
function is always increasing and to the right of ,
again staying in I, we are always
decreasing. In this case we can say that
the absolute maximum of in I
will occur at .
Similarly, if we know that to the left of the function is always decreasing and to the
right of the function is always increasing then the
absolute minimum of in I
will occur at .
Before we give a summary of this method let’s discuss the
continuity requirement a little. Nowhere
in the above discussion did the continuity requirement apparently come into
play. We require that that the function
we’re optimizing to be continuous in I
to prevent the following situation.
In this case, a relative maximum of the function clearly
occurs at . Also, the function is always decreasing to
the right and is always increasing to the left.
However, because of the discontinuity at ,
we can clearly see that and so the absolute maximum of the function
does not occur at . Had the discontinuity at not been there this would not have happened
and the absolute maximum would have occurred at .
Here is a summary of this method.
First Derivative Test
for Absolute Extrema