Because these notes are also being presented on the web
we’ve broken the optimization examples up into several sections to keep the
load times to a minimum. Do not forget
the various methods for
verifying that we have the optimal value that we looked at in the previous
section. In this section we’ll just use
them without acknowledging so make sure you understand them and can use them. So let’s get going on some more examples.
Example 1 A
window is being built and the bottom is a rectangle and the top is a
semicircle. If there is 12 meters of
framing materials what must the dimensions of the window be to let in the
most light?
Solution
Okay, let’s ask this question again in slightly easier to
understand terms. We want a window in
the shape described above to have a maximum area (and hence let in the most
light) and have a perimeter of 12 m (because we have 12 m of framing
material). Little bit easier to
understand in those terms.
Here’s a sketch of the window. The height of the rectangular portion is h and because the semicircle is on top
we can think of the width of the rectangular portion at 2r.
The perimeter (our constraint) is the lengths of the three
sides on the rectangular portion plus half the circumference of a circle of
radius r. The area (what we want to maximize) is the
area of the rectangle plus half the area of a circle of radius r.
Here are the equations we’ll be working with in this example.
In this case
we’ll solve the constraint for h
and plug that into the area equation.
The first and
second derivatives are,
We can see that
the only critical point is,
We can also see
that the second derivative is always negative (in fact it’s a constant) and
so we can see that the maximum area must occur at this point. So, for the maximum area the semicircle on
top must have a radius of 1.6803 and the rectangle must have the dimensions
3.3606 x 1.6803 (h x 2r).

Example 2 Determine
the area of the largest rectangle that can be inscribed in a circle of radius
4.
Solution
Huh? This problem
is best described with a sketch. Here
is what we’re looking for.
We want the area of the largest rectangle that we can fit
inside a circle and have all of its corners touching the circle.
To do this problem it’s easiest to assume that the circle
(and hence the rectangle) is centered at the origin. Doing this we know that the equation of the
circle will be
and that the right upper corner of the rectangle will have
the coordinates . This means that the width of the rectangle
will be 2x and the height of the
rectangle will be 2y. The area of the rectangle will then be,
So, we’ve got the function we want to maximize (the area),
but what is the constraint? Well since
the coordinates of the upper right corner must be on the circle we know that x and y must satisfy the equation of the circle. In other words, the equation of the circle
is the constraint.
The first thing to do then is to solve the constraint for
one of the variables.
Since the point that we’re looking at is in the first
quadrant we know that y must be
positive and so we can take the “+” part of this. Plugging this into the area and computing
the first derivative gives,
Before getting the critical points let’s notice that we
can limit x to the range since we are assuming that x is in the first quadrant and must
stay inside the circle. Now the four
critical points we get (two from the numerator and two from the denominator)
are,
We only want critical points that are in the range of
possible optimal values so that means that we have two critical points to
deal with : and . Notice however that the second critical
point is also one of the endpoints of our interval.
Now, area function is continuous and we have an interval
of possible solution with finite endpoints so,
So, we can see that we’ll get the maximum area if and the corresponding value of y is,
It looks like
the maximum area will be found if the inscribed rectangle is in fact a
square.

We need to again make a point that was made several times in
the previous section. We excluded
several critical points in the work above.
Do not always expect to do that.
There will often be physical reasons to exclude zero and/or negative
critical points, however, there will be problems where these are perfectly
acceptable values. You should always
write down every possible critical point and then exclude any that can’t be
possible solutions. This keeps you in
the habit of finding all the critical points and then deciding which ones you
actually need and that in turn will make it less likely that you’ll miss one
when it is actually needed.
Example 3 Determine
the point(s) on that are closest to (0,2).
Solution
Here’s a quick sketch of the situation.
So, we’re looking for the shortest length of the dashed
line. Notice as well that if the
shortest distance isn’t at there will be two points on the graph, as
we’ve shown above, that will give the shortest distance. This is because the parabola is symmetric
to the yaxis and the point in
question is on the yaxis. This won’t always be the case of course so
don’t always expect two points in these kinds of problems.
In this case we need to minimize the distance between the
point (0,2) and any point that is one the graph (x,y). Or,
If you think about the situation here it makes sense that
the point that minimizes the distance will also minimize the square of the
distance and so since it will be easier to work with we will use the square
of the distance and minimize that. If
you aren’t convinced of this we’ll take a closer look at this after this
problem. So, the function that we’re
going to minimize is,
The constraint in this case is the function itself since
the point must lie on the graph of the function.
At this point there are two methods for proceeding. One of which will require significantly
more work than the other. Let’s take a
look at both of them.
Solution 1
In this case we will use the constraint in probably the
most obvious way. We already have the
constraint solved for y so let’s
plug that into the square of the distance and get the derivatives.
So, it looks like there are three critical points for the
square of the distance and notice that this time, unlike pretty much every
previous example we’ve worked, we can’t exclude zero or negative
numbers. They are perfectly valid
possible optimal values this time.
Before going any farther, let’s check these in the second
derivative to see if they are all relative minimums.
So, is a relative maximum and so can’t possibly
be the minimum distance. That means
that we’ve got two critical points. The question is how do we verify that
these give the minimum distance and yes we did mean to say that both will
give the minimum distance. Recall from
our sketch above that if x gives
the minimum distance then so will x and so if gives the minimum distance
then the other should as well.
None of the methods we discussed in the previous section
will really work here. We don’t have
an interval of possible solutions with finite endpoints and both the first
and second derivative change sign. In
this case however, we can still verify that they are the points that give the
minimum distance.
First, notice that if we are working on the interval then the endpoints of this interval (which
are also the critical points) are in fact where the absolute minimum of the
function occurs in this interval.
Next we can see that if then . Or in other words, if the function is decreasing until it hits and so must always be larger than the
function at .
Similarly, then and so the function is always increasing to
the right of and so must be larger than the function at .
So, putting all of this together tells us that we do in
fact have an absolute minimum at .
All that we need to do is to find the value of y for these points.
So, the points on the graph that are closest to (0,2) are,
Solution 2
The first solution that we worked was actually the long
solution. There is a much shorter
solution to this problem. Instead of
plugging y into the square of the
distance let’s plug in x. From the constraint we get,
and notice that the only place x show up in the square of the distance it shows up as x^{2} and let’s just plug this
into the square of the distance. Doing
this gives,
There is now a single critical point, ,
and since the second derivative is always positive we know that this point
must give the absolute minimum. So all
that we need to do at this point is find the value(s) of x that go with this value of y.
The points are then,
So, for significantly less work we got exactly the same
answer.

This previous example had a couple of nice points. First, as pointed out in the problem, we
couldn’t exclude zero or negative critical points this time as we’ve done in
all the previous examples. Again, be
careful to not get into the habit of always excluding them as we do many of the
examples we’ll work.
Next, some of these problems will have multiple solution
methods and sometimes one will be significantly easier than the other. The method you use is up to you and often the
difficulty of any particular method is dependent upon the person doing the
problem. One person may find one way
easier and other person may find a different method easier.
Finally, as we saw in the first solution method sometimes
we’ll need to use a combination of the optimal value verification methods we
discussed in the previous section.
Now, before we move onto the next example let’s take a look
at the claim above that we could find the location of the point that minimizes
the distance by finding the point that minimizes the square of the
distance. We’ll generalize things a
little bit,
Fact
This is simple enough to prove so let’s do that here. First let’s take the derivative of and see what we can determine about the
critical points of .
Let’s plug into this to get,
By assumption we know that exists and and therefore the denominator of this will
always exist and will never be zero.
We’ll need this in several places so we can’t forget this.
If then because we know that the denominator will
not be zero here we must also have . Likewise, if then we must have .
So, and will have the same critical points in which
the derivatives will be zero.
Next, if doesn’t exist then will also not exist and likewise if doesn’t exist then because we know that the
denominator will not be zero then this means that will also not exist. Therefore, and will have the same critical points in which
the derivatives does not exist.
So, the upshot of all this is that and will have the same critical points.
Next, let’s notice that because we know that then and will have the same sign and so if we apply the
first derivative test (and recalling that they have the same critical points)
to each of these functions we can see that the results will be the same and so
the relative extrema will occur at the same points.
Note that we could also use the second derivative test to
verify that the critical points will have the same classification if we wanted
to. The second derivative is (and you
should see if you can use the quotient rule to verify this),
Then if is a critical point such that (and so we can use the second derivative test)
we get,
Now, because we know that and by assumption we can see that and will have the same sign and so will have the
same conclusion from the second derivative test.
So, now that we have that out of the way let’s work some
more examples.
Example 4 A
2 feet piece of wire is cut into two pieces and one piece is bent into a
square and the other is bent into an equilateral triangle. Where should the wire cut so that the total
area enclosed by both is minimum and maximum?
Solution
Before starting the solution recall that an equilateral
triangle is a triangle with three equal sides and each of the interior angles
are (or ).
Now, this is another problem where the constraint isn’t
really going to be given by an equation, it is simply that there is 2 ft of
wire to work with and this will be taken into account in our work.
So, let’s cut the wire into two pieces. The first piece will have length x which we’ll bend into a square and
each side will have length . The second piece will then have length (we just used the constraint here…) and
we’ll bend this into an equilateral triangle and each side will have length . Here is a sketch of all this.
As noted in the sketch above we also will need the height
of the triangle. This is easy to get
if you realize that the dashed line divides the equilateral triangle into two
other triangles. Let’s look at the
right one. The hypotenuse is while the lower right angle is . Finally the height is then the opposite
side to the lower right angle so using basic right triangle trig we arrive at
the height of the triangle as follows.
So, the total area of both objects is then,
Here’s the
first derivative of the area.
Setting this
equal to zero and solving gives the single critical point of,
Now, let’s
notice that the problem statement asked for both the minimum and maximum
enclosed area and we got a single critical point. This clearly can’t be the answer to both,
but this is not the problem that it might seem to be.
Let’s notice
that x must be in the range and since the area function is continuous we
use the basic process for finding absolute extrema of a function.
So, it looks
like the minimum area will arise if we take while the maximum area will arise if we take
the whole piece of wire and bend it into a square.

As the previous problem illustrated we can’t get too locked
into the answers always occurring at the critical points as they have to this
point. That will often happen, but one
of the extrema in the previous problem was at an endpoint and that will happen
on occasion.
Example 5 A
piece of pipe is being carried down a hallway that is 10 feet wide. At the end of the hallway the there is a
rightangled turn and the hallway narrows down to 8 feet wide. What is the longest pipe that can be carried
(always keeping it horizontal) around the turn in the hallway?
Solution
Let’s start off with a sketch of the situation so we can
get a grip on what’s going on and how we’re going to have to go about solving
this.
The largest pipe that can go around the turn will do so in
the position shown above. One end will
be touching the outer wall of the hall way at A and C and the pipe
will touch the inner corner at B. Let’s assume that the length of the pipe in
the small hallway is while is the length of the pipe in the large
hallway. The pipe then has a length of
.
Now, if then the pipe is completely in the wider
hallway and we can see that as then . Likewise, if the pipe is completely in the narrow hallway
and as we also have . So, somewhere in the interval is an angle that will minimize L and oddly enough that is the length
that we’re after. The largest pipe
that will fit around the turn will in fact be the minimum value of L.
The constraint for this problem is not so obvious and
there are actually two of them. The
constraints for this problem are the widths of the hallways. We’ll use these to get an equation for L in terms of and then we’ll minimize this new equation.
So, using basic right triangle trig we can see that,
So, differentiating
L gives,
Setting this
equal to zero and solving gives,
Solving for gives,
So, if radians then the pipe will have a minimum
length and will just fit around the turn.
Anything larger will not fit around the turn and so the largest pipe
that can be carried around the turn is,

Example 6 Two
poles, one 6 meters tall and one 15 meters tall, are 20 meters apart. A length of wire is attached to the top of
each pole and it is also staked to the ground somewhere between the two
poles. Where should the wire be staked
so that the minimum amount of wire is used?
Solution
As always let’s start off with a sketch of this situation.
The total length of the wire is and we need to determine the value of x that will minimize this. The constraint in this problem is that the
poles must be 20 meters apart and that x
must be in the range . The first thing that we’ll need to do here
is to get the length of wire in terms of x,
which is fairly simple to do using the Pythagorean Theorem.
Not the nicest function
we’ve had to work with but there it is.
Note however, that it is a continuous function and we’ve got an
interval with finite endpoints and so finding the absolute minimum won’t
require much more work than just getting the critical points of this function. So, let’s do that. Here’s the derivative.
Setting this
equal to zero gives,
It’s probably
been quite a while since you’ve been asked to solve something like this. To solve this we’ll need to square both
sides to get rid of the roots, but this will cause problems as well soon
see. Let’s first just square both
sides and solve that equation.
Note that if
you can’t do that factoring don't worry, you can always just use the quadratic
formula and you’ll get the same answers.
Okay two issues
that we need to discuss briefly here.
The first solution above (note that I didn’t call it a critical
point…) doesn’t make any sense because it is negative and outside of the
range of possible solutions and so we can ignore it.
Secondly, and
maybe more importantly, if you were to plug into the derivative you would not get zero
and so is not even a critical point.
How is this possible? It is a solution
after all. We’ll recall that we
squared both sides of the equation above and it was mentioned at the time
that this would cause problems. We’ll
we’ve hit those problems. In squaring
both sides we’ve inadvertently introduced a new solution to the
equation. When you do something like
this you should ALWAYS go back and verify that the solutions that you get are in
fact solutions to the original equation.
In this case we were lucky and the “bad” solution also happened to be
outside the interval of solutions we were interested in but that won’t always
be the case.
So, if we go
back and do a quick verification we can in fact see that the only critical
point is and this is nicely in our range of
acceptable solutions.
Now all that we
need to do is plug this critical point and the endpoints of the wire into the
length formula and identify the one that gives the minimum value.
So, we will get
the minimum length of wire if we stake it to the ground feet from the smaller pole.

Let’s do a modification of the above problem that asks a
completely different question.
Example 7 Two
poles, one 6 meters tall and one 15 meters tall, are 20 meters apart. A length of wire is attached to the top of
each pole and it is also staked to the ground somewhere between the two
poles. Where should the wire be staked
so that the angle formed by the two pieces of wire at the stake is a maximum?
Solution
Here’s a sketch for this example.
The equation that we’re going to need to work with here is
not obvious. Let’s start with the
following fact.
Note that we need to make sure that the equation is equal
to because of how we’re going to work this problem. Now, basic right triangle trig tells us the
following,
Plugging these
into the equation above and solving for gives,
Note that this
is the reason for the in our equation. The inverse tangents give angles that are
in radians and so can’t use the 180 that we’re used to in this kind of
equation.
Next we’ll need
the derivative so hopefully you’ll recall how to differentiate
inverse tangents.
Setting this
equal to zero and solving give the following two critical points.
The first critical point is not in the interval of
possible solutions and so we can exclude it.
It’s not difficult to show that if that and if that and so when we will get the maximum value of .

Example 8 A
trough for holding water is be formed by taking a piece of sheet metal 60 cm
wide and folding the 20 cm on either end up as shown below. Determine the angle that will maximize the amount of water that
the trough can hold.
Solution
Now, in this case we are being asked to maximize the
volume that a trough can hold, but if you think about it the volume of a
trough in this shape is nothing more than the crosssectional area times the
length of the trough. So for a given
length in order to maximize the volume all you really need to do is maximize
the crosssectional area.
To get a formula for the crosssectional area let’s redo
the sketch above a little.
We can think of the crosssectional area as a rectangle in
the middle with width 20 and height h
and two identical triangles on either end with height h, base b and
hypotenuse 20. Also note that basic
geometry tells us that the angle between the hypotenuse and the base must
also be the same angle that we had in our original sketch.
Also, basic right triangle trig tells us that the base and
height can be written as,
The
crosssectional area for the whole trough, in terms of ,
is then,
The derivative of the area is,
So, we have either,
However, we can
see that must be in the interval or we won’t get a trough in the proper
shape. Therefore, the second critical
point makes no sense and also note that we don’t need to add on the standard
“ ” for the same reason.
Finally, since
the equation for the area is continuous all we need to do is plug in the
critical point and the end points to find the one that gives the maximum
area.
So, we will get
a maximum crosssectional area, and hence a maximum volume, when .
