We will continue working on optimization. Please complete the following
optimization problem on your own.
Example 0.1. 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?
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.
Hopefully that problem was within reason. Remember, you need to begin by
identifying the quantity you want to maximize or minimize, writing this quantity
as a function of a single variable, finding the critical points, and plugging the
critical points into the original function.
Let’s do another simple example:
Example 0.2. There is a buoy out at sea along a straight coastline. The buoy
is 100 meters away from the nearest point on the shore, and you are currently
200 meters away from the nearest point on the shore to the buoy. You can run
at a rate of 3m/s and swim at a rate of 1m/s. How far should you run along
the shore before swimming out to the buoy in order to reach the buoy as quickly
as possible?
We should start by identifying the quantity we want to maximize or minimize
(we want to minimize the total time t that it takes to reach the buoy) by
manipulating the distance d you run along the shoreline.
Let t1 be the running time and t2 be the swimming time. Then the total
time t = t1 + t2 . We can compute t1 and t2 in the following way: the running
distance, d, satisfies d = 3t1 , so t1 = d3 . Now what about t2 ? Well, we need
to find the total swimming distance s. We can see by using the Pythagorean
1
p
theorem that
s2 = (200 − d)2 + 1002 , so s = (200 − d)2 + 1002 . Therefore, we
√
2
2
get t2 = 200−d1 +100 . So the total time t can be expressed in terms of d as
t=
d p
+ 50000 − 400d + d2 .
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Now, we can compute the derivative of t:
t′ =
1 1
+ (50000 − 400d + d2 )−1/2 (2d − 400)
3 2
which simplifies to
1
d − 200
+
.
3 (50000 − 400d + d2 )1/2
This is equal to zero if
d − 200
−1
=
2
1/2
3
(50000 − 400d + d )
which happens when
600 − 3d = (50000 − 400d + d2 )1/2 .
Square both sides:
360000 − 3600d + 9d2 = 50000 − 400d + d2
and put everything on one side:
8d2 − 3200d + 310000 = 0
This is solved by
d=
3200 ±
√
32002 − 4 ∗ 310000 ∗ 8
16
which is
d=
3200 ±
which is
200 ±
Notice that 200 +
√
320000
16
√
1250.
√
1250 isn’t in the domain, so we want
√
200 − 1250,
which is about 165 meters. Plugging into the original function gives a value
close to 160 seconds.
Plugging in the extreme value d = 0 gives a time of about 223, and plugging
in the extreme√value 200 gives 166.67 seconds. So the time is minimized when
you run 200 − 1250 meters along the shore and then swim the rest of the way.
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