Today, and for the next two weeks, we will discuss the topic of optimization.
In optimization problems, the goal is to minimize or maximize a given function,
typically on a bounded interval.
The first thing to do in these kinds of problems is to identify which quantity
you’re trying to maximize or minimize. In the language of optimization this
is typically called the objective function. We also need to identify the variable
with respect to which we are optimizing the objective function. Then, we use
the strategy discussed last week to locate the global maximum or minimum.
Let’s start with a simple example.
Example 0.1. Given all isosceles triangles of length 10, which one has the
largest area?
To tackle this problem, we must first identify the quantity to be maximizedthe area- and then identify the quantity that we have control over- in this case
the base length (or the leg length, or, more inconveniently, the base angle or
the angle between the legs). The base is the easiest quantity to deal with.
Notice that the length, b, of the base is related to the perimeter by the
equation
b + 2ℓ = p
where p is the perimeter, b is the base, and ℓ is the length of each leg. Therefore,
10−b
we get that each of the legs has length p−b
in light of the fact
2 , which is
2
that the triangles under consideration have perimeter equal to 10.
Now, drop a perpendicular onto the base of the triangle. This subdivides
the original triangle into 2 right triangles. One of the legs of the right triangle
has length b/2, and the hypotenuse has length 5 − 2b . Therefore, the other leg
has length
s
2 2
b
b
−
5−
2
2
which comes out to
√
25 − 5b.
So the total area is
b√
25 − 5b.
2
Now, we have achieved our goal of writing the area in terms of b. Which values
of b make sense? Clearly b ≥ 0, and also we need b ≤ 5 because the sum of 2 side
lengths is at least as long as the third side. The inequalities are strict provided
that the triangle under consideration is nondegenerate, and we see that plugging
in b = 0 or b = 5 gives an area of zero.
So we differentiate:
!
− 5b
1
′
1/2
2
A =
(25 − 5b) +
2
(25 − 5b)1/2
1
This is the same as
A′ =
50 − 15b
4(25 − 5b)1/2
which is zero when b = 10
3
Now, we need to test our critical points. Plugging in x√= 0 and x = 5 gives
25 3
an area of zero, and plugging in x = 10
3 gives an area of
9 , which is positive.
Another common type of example involves shapes where a side is not present:
Example 0.2. An open-topped box with a square base needs to be made to hold
100cm3 of water. What should the dimensions of the box be in order to minimize
the surface area?
This time, we want to minimize the surface area of the box, given that the
volume is 100cm3. If s is the side length of the box, we have that the surface
area is
A = 4sh + s2
and the volume is
100 = s2 h
in other words, we get that
h=
so that means that we get
A=
100
s2
400
+ s2 .
s
We differentiate to get
400
+ 2s
s
and we set the derivative equal to zero to get
A′ = −
0=−
400
+ 2s
s
400
= 2s
s
200 = s2
√
s = 10 2
Now, there is no upper bound on the size, s, but we notice that the surface
area gets very√large as s → ∞. The same thing happens as s → 0. When we
40
plug in s = 10 2, we get √
+ 200, which is therefore the minimal surface area.
2
Now, you should attempt this easier example problem on your own:
Example 0.3. A farmer is putting up a pen to contain his sheep. The bounded
area is to be a rectangle, one side of which is on a river (and therefore doesn’t
need any fencing). The farmer has only 100 meters of fencing. What dimensions
should the rectangle be?
2
Let s be the short side of the rectangle and ℓ the long side. Then we have
100 = 2s + ℓ
and therefore that
ℓ = 100 − 2s.
Therefore, the area is
ℓs = (100 − 2s)s
which is equal to
100s − 2s2
and when we differentiate we get
A′ = 100 − 4s
setting A′ = 0 gives s = 25.
You should imagine that s could be as small as zero (giving zero area) or as
large as 50 (again giving zero area), so the critical point s = 25 maximizes the
area, giving an area of 50 ∗ 25 = 1250 square meters.
3