Math 181 - Practice Exam 2 - solutions
Problem 1 A population of dinosaurs produces offspring which numbers
P (t) = −0.3t2 +0.1t+10 for times t in the interval [−4, 0]. At the same time,
dinosaurs die from various causes, deaths numbering D(t) = 0.2t2 + 0.1t in
the same timespan.
a)The net increase of the population is offspring - death numbers. Find the
rate of change of this net increase at t = −4.
b) Find the equation of the tangent line of P (t) at t = −4 and sketch it,
together with the graph of P , over the interval [−4, 0]. Your graph should
include the values of P at −4, −2, 0.
a) Net increase is
P (t) − D(t) = −0.5t2 + 10
and this has derivative Dt (P − D) = −t, so the rate of change at t = −4 is
4.
b) Note that this part talks about P , not P − D. The derivative here is
P 0 (t) = −0.6t + 0.1, and therefore P 0 (−4) = 2.5. We also find P (−4) = 4.8,
so the tangent line there has equation
y = 2.5(t + 4) + 4.8
Further values are P (−2) = 8.6 and P (0) = 10.
20
18
16
14
12
10
8
6
4
2
0
ï6
ï2
ï4
0
2
t
Graph of the function from Problem 1
10.0
7.5
5.0
2.5
0.0
ï4
ï2
t
0
2
ï2.5
ï5.0
ï7.5
Graph of the function from Problem 2
Problem 2 Consider the function
2t − 3
f (t) = 2
t + 2t − 4
a) Find the derivative of f using quick rules.
b) Find all relative maxima and minima of f .
c) Tell where the function f is increasing and where it is decreasing.
d) Find the maximum and minimum value of f on the interval [−3, 0].
e) Find the equation of the tangent line to the graph of f at t = 2.
For a),
−2t2 + 6t − 2
2(t2 + 2t − 4) − (2t − 3)(2t + 2)
=
,
(t2 + 2t − 4)2
(t2 + 2t − 4)2
√
b) There are singular points at t = −1 ± 5 where the denominator is zero.
These numbers are approximately −3.24 and 1.24. Solving f 0 (t) = 0 for
stationary points,
√
−6 ± 36 − 16
3 1√
t=
= ±
5.
−4
2 2
f 0 (t) =
These numbers are approximately 2.62 and 0.38.
This gives the following sign pattern.
t
−3.24
0.38
1.24
2.62
sign of f 0 (t) −
|
−
+
|
+
−
According to the sign pattern, 0.38 is a relative minimum and 2.6 is a relative
maximum. At the other two critical points, f (t) is not even defined, so they
are not relative extrema.
d) All stationary points are outside [−3, 0]! So we only need to look at the
endpoints -3, 0.
t −3 0
f (t) 9 0.75
So the maximum value is 9 (attained at t = −3) and the minimum value is
0.75 (attained at t = 0).
e) At t = 2, we have f (2) = 1/4 and f 0 (2) = 1/8, so the tangent line has an
equation
1
1
1
y = (t − 2) + = t
8
4
8
which means it actually passes through the origin (pure chance).
Problem 3 A farmer wants to fence in a rectangular field using strong
fence on two opposite sides, costing $ 5 per foot, and cheap fence on the two
other sides, costing $ 4 per foot.
a) What is the minimum cost to fence in 4000 square feet in this manner?
b) What if the farmer can use an existing 60-foot wall instead of strong
fence?
a) Write a for the length of one side with strong fence, b for the length of one
side with cheap fence. So the area is A = ab = 4000, leading to b = 4000/a.
Then consider the cost C of a field with these dimensions. We need 2a feet
of strong fence and 2b of the cheap, costing 10a and 8b (dollars), respectively.
The total cost is
C = 10a + 8b = 10a +
32, 000
a
so we have written C = C(a) as a function of a! Now we want to minimize
C(a). Look for critical points: Obviously, we only have to investigate a > 0,
so no singular points. Then C 0 (a) = 0 means
0 = 10 −
32000
a2
a2 = 3200
√
a =
3200 ≈ 56.57
(in feet). The negative value for a is discareded, and we get a minimum cost
of
√
√
√
C( 3200) = 10 3200 + 32000/ 3200 ≈ 1131.4
(in dollars). Yes, this really is the minimum and not a maximum, because
the sign pattern of the derivative is −+.
Note how the cheap fence and the strong fence both contribute the same
amount to the total!
b) For the cost, we need to look at two different scenarios. If we use all of
the existing wall (meaning a ≥ 60), the setup changes to
C = 5a + 5(a − 60) + 8b = 10a − 300 +
32, 000
a
which we differentiate to get the same derivative
32, 000
a2
√
0
as before!
In
particular,
C
(a)
>
0
for
a
>
3200, so C(a) is increasing for
√
a > 3200. This means for a ≥ 60, the lowest-cost fence is actually built
with a = 60.
The second scenario is that we don’t use all of the existing wall (so a ≤
60). As the cost, we get
C 0 (a) = 10 −
32, 000
a
√
with C 0 (a) = 5 − 32, 000/a2 equal to zero for a = 6400 = 80 only. So inside
the interval (0, 60), always C 0 (a) < 0. This means the lowest-cost fence is
built using a = 60 – and that is the final answer.
C = 5a + 8b = 5a +
Problem 4 Find the following derivatives, using Quick Rules. No need
to simplify.
√
3x2 − 4x + 2
x+3
y = 2
x −4
y = (x2 − 1)2
y = (x3 + 2x − 1)100
y =
√
Dx ( 3x2 − 4x + 2)
x+3
Dx
x2 − 4
Dx (x2 − 1)2
Dx x3 + 2x − 1)100
√
= 2 3x − 4
x2 − 4 − 2x(x + 3)
(Q.R.)
=
(x2 − 4)2
= 4x(x2 − 1)
(P.R.)
2
3
= 100(3x + 2)(x + 2x − 1)99
Problem 5 A cylindrical barrel is to be built from sheet metal. Assume
that the surface area of the barrel is
A = 2πr2 + 2πhr
if h is its height and r the radius of the base. The volume is V = πr2 h. If
you want to build a barrel with a volume of 432π cubic feet, what are the
dimensions that use the least amount of material?
From the volume equation, we get
432
V
h= 2 = 2 .
πr
r
Using this to express A as a function of r alone,
2 · 432π
A = 2πr2 +
r
The amount of material we need is proportional to A, so we are seeking to
minimize A. For critical numbers, we study
2 · 432π
A0 (r) = 4πr −
r2
√
which we solve for r, r = 3 216 = 6. The sign pattern of V 0 (r) is −+ which
means r = 6 really gives the minimum (it follows that h = 12, from the
volume equation).