Math 19: Group Worksheet 2 Solutions
1. A box with a square base and an open top must have a volume of 32,000 cm3 . Find the
dimensions of the box that minimize the amount of material used.
We begin by drawing a picture.
h
x
x
Our goal is to minimize surface area, A, which is given by A = x2 + 4xh. We need to
eliminate a variable, either x or h; I’ll eliminate h. We use the volume, V = x2 h = 32000,
so h = 32000/x2 . This yields
128000
32000
= x2 +
,
A = x2 + 4x ·
2
x
x
and the domain is (0, ∞). We find that
128000
2x3 − 128000
2(x3 − 64000)
=
=
,
x2
x2
x2
√
so x = 3 64000 = 40 is the only critical point; the corresponding h is 20. This is probably
our answer, but we must check that it actually is! There are at least two ways to do
this.
First way: Look at A0 on the number line:
−
+
A0
0
3
− 40
+
x − 64000
2
+
+
x
A0 = 2x −
Thus, we see that x = 40 is a local minimum, and in fact must be a global/absolute minimum. (You don’t need to have factored A0 to do this, but it makes it easier.)
Second way: Consider the extreme values of the domain, x = 0 and x = ∞. Neither of
these points are in the domain, but we can compute the limit; we find
128000
2
= +∞ and lim A = +∞.
lim A = lim+ x +
x→∞
x→0+
x→0
x
Since x = 40 is the only critical point, and the function blows up at the endpoints, x = 40
must give the minimum value.
2. A boat leaves a dock at 2:00 PM and heads due south at a speed of 20 km/h. Another
boat has been traveling due east at 15 km/h and reaches the same dock at 3:00 PM. At
what time were the two boats closest together?
Call the first boat A and the second B, and suppose that t hours have past since 2 PM;
note that our domain is 0 ≤ t ≤ 1. We draw a picture.
B at 2 PM
B at time t
x
b
Dock
y
D
A at time t
Our goal is to minimize D, which we can find via the Pythogorean theorem: D2 = x2 + y 2 .
We thus need to express x and y in terms of t, for which our main tool is Distance = Rate·
Time. Using the rates of boats A and B, we find that y = 20t and that b = 15t. We need
to use the latter to find x. Since B reaches the dock at 3 PM after traveling for an hour at
15 km/h, at 2 PM it is 15 km west of the dock. Thus, b + x = 15, so x = 15 − 15t. To recap,
this yields
D2 = x2 + y 2 = (15 − 15t)2 + (20t)2 = 225t2 − 450t + 225 + 400t2
= 625t2 − 450t + 225.
Our goal is to minimize D, but this is the same as minimizing D2 . (This is a useful trick
which saves some annoying algebra, and is often used in distance problems.) We find that
dD2
= 1250t − 450,
dt
so the only critical point is at t = 450/1250 = 9/25. Once again, this is probably our
answer, but we must check!
2
First way: Use a number line for dD
:
dt
dD2
dt
0
1250t − 450
−
− 9/25
+
1
+
We see that D2 is decreasing down to t = 9/25 and then increasing after, so t = 9/25 is
the minimum.
Second way: Plug in endpoints and CP’s. We find that
s
2
9
9
D|t=0 = 15, D|t=1 = 20, and D|t=9/25 = 625
− 450
+ 225
25
25
√
= 144 = 12.
Thus, t = 9/25 yields the minimum value.
3. A cone-shaped drinking cup is made from a circular piece of paper of radius R by cutting out a sector and the joining the newly formed edges. Find the maximum capacity of
such a cup.
A
O
h
O
r A
(Notice the additions to the picture of the cone.)
We’re trying to maximize the volume V of the cone, where V = 13 πr2 h, with r and h as in
the picture. This is too many variables, so we need to eliminate one. Since OA = R, we
have that r2 + h2 = R2 , and so r2 = R2 − h2 . (Notice that our formula for V has an r2 in
it, so we don’t actually need to solve for r. If we eliminate h instead, we’d have something
involving square roots, which is ugly, but works.)
Thus, we can write our volume as
1
1
1
V = π(R2 − h2 )h = πR2 h − πh3 .
3
3
3
The domain is 0 ≤ h ≤ R. (The former case corresponds to not removing anything from
the circle, the latter to removing all but a sliver.)
We find that
1
V 0 = πR2 − πh2 ,
3
q
1
πR2
which means that our critical points are h = ± 3 π = ± √R3 . But since the domain is
0 ≤ h ≤ R, only the point h = √R3 is in our domain. We find it convenient to check
whether this is the maximum by plugging it and the endpoints into our function V ,
though it is also possible to do it with the number line.
In particular, we find that V |h=0 = 0 and V |h=R = 0, while
1
R2
R
2π
2
√
√
V |h=R/ 3 = π R −
= √ R3 > 0,
3
3
3
9 3
so our CP is indeed the maximum.
4. If a snowball melts so that its surface area decreases at a rate of 1 cm2 /min, find the
rate at which the diameter decreases when it is 10 cm.
This is a related rates problem, not optimization. The surface area of a sphere is
2
D
2
= πD2 ,
A = 4πr = 4π
2
where D is the diameter. We’re told that dA
= −1 cm2 /min, and we’re trying to find
dt
when D = 10 cm. Using the above formula, we have that
dD
dA
= 2πD
,
dt
dt
and plugging in our values, we find that
dD
−1 cm2 /min = 2π(10 cm)
,
dt
−1
so that dD
= 20π
cm/min.
dt
(More on next page.)
dD
dt
5. An oil refinery is located on the north bank of a straight river that is 2 km wide. A
pipeline is to be constructed from the refinery to storage tanks located on the south bank
of the river 6 km east of the refinery. The cost of laying pipe overland to a point P on the
north bank is $600,000/km, and the cost of laying pipe under water is $800,000/km. To
minimize the cost of the pipeline, where should the point P be located?
As before, we begin with a picture.
Refinery
6−x
P
x
2 km
`
Tanks
6 km
We’re trying to minimize the cost C, which, in hundreds of thousands of dollars, is given
by
C = (6 − x) · 6/km + ` · 8/km.
√
Using the Pythagorean theorem, we find that ` = x2 + 4, so
√
C = 36 − 6x + 8 x2 + 4.
The domain is technically (−∞, ∞), but it is apparent we may restrict to [0, 6]. We find
that
√
8 · 2x
−6 x2 + 4
8x
0
C = −6 + √
= √
+√
2 x2 + 4
x2 + 4
x2 + 4
√
8x − 6 x2 + 4
√
=
.
x2 + 4
√
To find the critical points, we set 8x − 6 x2 + 4 = 0, and find
√
8x = 6 x2 + 4
64x2 = 36(x2 + 4)
28x2 = 4 · 36
36
x2 = ,
7
is in our domain. We plug this and our endpoints into
so x = ± √67 , of which only x = √67
C:
√
√
C|x=0 = 52, C|x=6 = 8 40 ≈ 50.59, and C|x= √6 = 36 + 4 7 ≈ 46.58.
7
Thus, x =
√6
7
is the minimum.
6. Evaluate each of the following limits, using L’Hôpital’s rule if necessary.
ex
a. lim 2
x→∞ x
, so we can use
“Plugging in” x = ∞, we see that the limit is of the form ∞
∞
L’Hôpital’s rule. We find
ex
ex
,
lim 2 = lim
x→∞ x
x→∞ 2x
which is again of the form ∞
. We once again apply L’Hôpital, which yields
∞
ex
lim
= +∞.
x→∞ 2
ex
x→−∞ x2
b. lim
Notice that lim ex = 0, so that this limit is of the form
Note:
x→−∞
0
is
Because ∞
0
,
∞
which is just 0.
not indeterminate, it is wrong to apply L’Hôpital’s rule here.
ex
x→0 x2
c. lim
This is of the form 10 , which is once again not indeterminate; thus, we cannot
apply L’Hôpital’s rule. Instead, this is some flavor of infinity, and since both ex
and x2 are always positive, it will be +∞.