At
t
= 9, therefore,
A
(9) = $100
e
0
.
45
.
034
10.0points
A software company finds that the marginal
price at
x
units of demand per week for its
software is proportional to the price
p
. At a
price of $100 there is no demand, while the
demand is for 70 units at a price of $10. How
should the company price the software if it
wishes to sell 90 units per week?
1.
price = $6
.
17
2.
price = $8
.
17
3.
price = $7
.
17
4.
price = $5
.
17
correct
5.
price = $9
.
17
Explanation:
As a function of demand
x
, the marginal
price is given by the differential equation
dp
dx
=
−
kp,
since the marginal price is proportional to the
price
p
; note that the derivative will be neg-
ative because demand will decrease as price

pacheco (jnp926) – Homework 5 – staron – (52840)
18
increases.
After Integration this differential
equation becomes
integraldisplay
1
A
dA
=
integraldisplay
r
100
dt.
Thus
ln
p
=
−
kx
+
C.
In turn, after exponentiation this becomes
p
(
x
) =
e
C
e
−
kx
=
Ae
−
kx
with
C
, and hence
A
, an arbitrary constant.
The value of
A
is detemined by the condition
p
(0) = 100, for then
p
(
x
) = 100
e
−
kx
.
On the other hand, the value of
k
is deter-
mined by the condition
p
(70) = 10, for then
10 = 100
e
−
70
k
, so
k
=
1
70
ln
parenleftBig
100
10
parenrightBig
.
Thus
p
(
x
) = 100
e
−
x
70
ln(100
/
10)
.
At
x
= 90, therefore,
p
(90) = 100
e
−
90
70
ln(10)
= $5
.
17
.
035
10.0points
Scientists began studying the elk popula-
tion in Yellowstone Park in 1990 when there
were 700 elk.
They determined that
t
years
after the study began the population size,
N
(
t
), was increasing at a rate proportional to
1300
−
N
(
t
). If the population was 800 in year
2000, estimate the size of the elk population
in year 2010?
1.
population size
≈
904
2.
population size
≈
944
3.
population size
≈
884
correct
4.
population size
≈
924
5.
population size
≈
864
Explanation:
The population size,
N
(
t
),
satisfies the
equations
dN
dt
=
k
(1300
−
N
)
,
N
(0) = 700
with
k
a constant. Thus
−
ln(1300
−
N
) =
kt
+
C
which after exponentiation becomes
N
(
t
) = 1300
−
Ae
−
kt
where
A
is an arbitrary constant.
Now
A
is
determined by the initial condition on
N
(
t
)
because
N
(0) = 700
=
⇒
A
= 600
.
Consequently,
N
(
t
) = 1300
−
600
e
−
kt
.
On the other hand, the value of
k
is deter-
mined by the population size in year 2000
when
N
(10) = 800, for then
500 = 600
e
−
10
k
.
Taking logs of both sides we thus see that
k
=
1
10
ln
parenleftBig
6
5
parenrightBig
.
In year 2010, therefore,
N
(20) = 1300
−
600
e
−
2ln
6
5
.
Hence, in year 2010,
population size
≈
884
.
036
10.0points

pacheco (jnp926) – Homework 5 – staron – (52840)
19
A population is modeled by the differential
equation
dP
dt
= 1
.
9
P
parenleftbigg
1
−
P
4000
parenrightbigg
.
For what values of
P
is the population in-
creasing?
1.
P >
2000
2.
0
< P <
4000
correct
3.
P >
4000
4.
P >
1
.
9
5.
0
< P <
1
.
9
Explanation:
1
.
7
P
parenleftBig
1
−
P
4500
parenrightBig
In order to check for increase, the derivative
must be above 0.
1
.
7
P
parenleftBig
1
−
P
4500
parenrightBig
>
0
1
.
7
P
−
1
.
7
P
2
4500
>
0
1
.
7
P >
1
.
7
P
2
4500
1
.
7
P
∗
4500
>
1
.
7
P
2
4500
∗
4500
7650
P
1
.
7
>
1
.
7
P
2
1
.
7
4500
P
P
>
P
2
P
4500
> P
.
P cannot be negative because it is used to
represent population.
037
10.0points
Scientists can determine the age of ancient ob-
jects by a method called
radiocarbondating
.