Last week, we discussed optimization problems where all of the numbers
were given. For example, we solved problems where we were given that the
surface area of, say, a box was 100cm2 and we had to determine the maximum
possible volume of the box.
This week, we will be considering problems like the following:
Example 1. An open-topped box of surface area A is going to be built with a
square base. Which dimensions should the box have to maximize the volume?
This is similar to a problem we’ve encountered before. However, the number
100cm2 was replaced by the parameter A.
The secret to this problem is that it’s not any different from the problem
where A was replaced by a number. Therefore, to solve the problem, we treat
A as if it were any old number, e.g. a constant. Our answer will depend on the
area A.
Let’s start by writing down the equations for the volume and the surface
area. If s is the side length of the base, and h is the height, we have
V = s2 h
and
A = 4sh + s2 .
As usual, we want to get V in terms of a single variable. In this case, it makes
sense to get v in terms of s, because h is easier to solve for in the equation for
A. So we solve for h:
A − s2
h=
4s
and we get
A
s
2
V =s
−
4s 4
So
As s3
− .
4
4
On the face of it, it looks like v depends on both A and s, and in fact it does.
However, for this problem A is a parameter - it’s fixed and we can’t change it. In
other words, it should be treated as if it were a constant number, like 100cm2 .
So when we differentiate, we treat A like a constant. We will now differentiate:
V =
V′ =
A 3s2
−
4
4
setting V ′ = 0 gives:
3s2 = A
so
s=
r
1
A
.
3
Note that this gives a volume of
A
V =
4
1/2
3/2
A
1 A
−
3
4 3
There is a power of A3/2 in both terms so we factor it out:
1
1
√ − √
V = A3/2
4 3 12 3
which is
3/2
V =A
2
√
12 3
,
a positive number.
Finally, we check
√ the endpoints: s can be as small as 0, giving a volume of
0, or as large as A, giving a volume of
A3/2
A3/2
−
4
4
q
which is zero. So we should select s to be A
3.
V =
Notice that the difficulty here is in determining which of the critical points
and endpoints gives the largest
q value. Here, it took some work to see that the
largest value you got was at A
3 - it wasn’t even immediately obvious that this
number was positive!
Sometimes, you’ll have a problem with more than one parameter in it:
Example 2. You have a piece of wire of length ℓ, and you are going to cut it.
You will bend one piece of the wire to make a regular polygon with n sides, and
bend the other piece of the wire to make a circle. How should you cut the wire
in order to maximize the area enclosed by both the polygon and the circle? The
area of a regular polygon with n sides is
Apolygon
s2 n
,
4 tan nπ
where s is the side length of the polygon.
We compute the total area
A = Apolygon + Acircle
and total perimeter
ℓ = ℓpolygon + ℓcircle
the perimeter of the polygon is
ℓpolygon = ns
2
and the circumference of the circle is
ℓ − ns
so the radius is
ℓ − ns
2π
and the area is
π
ℓ − ns
2π
2
.
Thus, we get the total area in terms of s:
s2 n
+π
A=
4 tan nπ
ℓ − ns
2π
2
Notice that A and s are the only variables here: n and ℓ are parameters. We’ll
expand the square so that it’s easier to deal with later:
ℓ2 − 2ℓns + n2 s2
s2 n
+
π
4π 2
4 tan πn
A=
which is
A=
s2 n
ℓ2 − 2ℓns + n2 s2
+
4π
4 tan πn
and now we differentiate. Remember s is the only variable here; we treat ℓ and
n as constants:
2sn2 − 2ℓn
2sn
+
A′ (s) =
4π
4 tan nπ
we set A′ = 0:
0=
2sn
2sn2 − 2ℓn
π +
4π
4 tan n
we can now divide by 2n since it occurs in every term:
0=
s
4 tan
π
n
This is a linear equation in s:
1
4 tan
ℓ
=
4π
so we get:
+
sn − ℓ
4π
n
π + 4π
n
!
ℓ
s=
4π
1
π
4 tan( n
)
3
+
n
4π
s
which comes out to
s=
ℓ
π
π
tan( n
)
+n
.
With this value of s, we get that we should cut the wire at length sn, which is
sn =
ℓ
π
π
n tan( n
)
+1
.
I’ll leave it to you guys to determine whether or not this is the maximum.
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