3. (Problem 1.5.36 from book.) A tank initially contains 60 gal of pure water. Salt water containing 1 lb
of salt per gallon enters the tank at 2 gal/min, and the perfectly mixed solution leaves the tank at 3
gal/min. Thus the tank is empty after 1 hour. Find the amount of salt in the tank after t minutes.
Determine the maximum amount of salt ever in the tank.
Solution: First note that the tank is losing a net of 1 gal/min of liquid, which means that it will be
empty in 60 minutes. More specifically, the number of gallons in the tank at time t is V (t) = 60 − t.
Second, we want to write down a differential equation for the salt at time t, which we will denote by
S(t). Clearly, the rate of change of salt will be the rate coming in minus the rate going out. The
amount coming in is 1 lb/ gal * 2 gal/min = 2 lb/min. The rate going out is more complicated, since
it depends on the concentration of the salt in the tank at the time. But notice that if there is S(t)
pounds of salt at time t, and there are 60 − t gallons in the tank at time t, then the concentration
is thus S(t)/(60 − t) lb/gallon in the tank. The rate of liquid leaving is 3 gal/min, so the rate out is
3S(t)/(60 − t). Thus we have
3S(t)
dS
=2−
,
dt
60 − t
or
3
S = 2.
S′ +
60 − t
We will use the integrating factor
e
R
3
60−t
dt
= e−3 ln(60−t) = eln((60−t)
−3
)
= (60 − t)−3 .
Then we have
(60 − t)−3 S ′ + 3(60 − t)−4 S = 2(60 − t)−3 ,
d
((60 − t)−3 S(t)) = 2(60 − t)−3 ,
dt
(60 − t)−3 S(t) = (60 − t)−2 + C,
S(t) = C(60 − t)3 + (60 − t).
Now, we also need an initial condition, but notice that there is no salt in the tank at the start, so that
S(0) = 0. Plugging in gives
S(0) = C ∗ 603 + 60 = 0,
or
1
1
=−
.
602
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Thus the solution for all t between 0 and 60 minutes is given by
C =−
S(t) = −
(60 − t)3
+ (60 − t).
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(We can check that S(60) = 0, as it should.) Now we want to know the maximum value of S(t) for t
between 0 and 60. We first compute its derivative
S ′ (t) =
−(60 − t)2
+ 1.
1200
Setting this equal to zero gives
(60 − t)2 = 1200
or
t = 60 ±
√
1200.
Since we only care about t less than 60, we want
to take the
√ solution
√ with the minus and not the plus.
3 √
Thus the value we are looking for is S(60 − 1200) = 20( 3 − 1/ 3) ≈ 23.094.
1.5.36. A tank initially contains 60 gallons of pure water. Brine containing 1 lb of salt per
gallon enters the tank at 2 gal/min, and the (perfectly mixed) solution leaves the tank at
3 gal/min; thus, the tank is empty after exactly 1 hour.
(a) Find the amount of salt in the tank after t minutes.
(b) What is the maximum amount of salt ever in the tank?
Solution (a): We produce a model for this system. Let x (t) denote the amount of salt in
the tank at time t, t in minutes since the brine starts flowing into the tank. The flow rate
in is ri = 2 gal/min, and the flow rate out is ro = 3 gal/min, so the volume in the tank at
time t is V (t) = 60 + (2 − 3)t = 60 − t, in gallons.
The concentration of salt in the tank (and hence of the solution flowing out of the tank) is
then co (t) = x (t)/V (t) = 601−t x (t). Since the incoming concentration is ci = 1 lb/gal, the
net rate of change of the salt is
dx
1
3
= ri ci − ro co = (2)(1) − (3)
x (t) = 2 −
x ( t ),
dt
60 − t
60 − t
which is a linear DE. Rearranging it so that xR(t) and x 0 (t) are on the left-hand side, we
have x 0 + 603−t x = 2. Then P(t) = 603−t , so P(t) dt = −3 ln |60 − t|, with the minus
sign coming from the chain rule with 60 − t. Hence, an integrating factor is e−3 ln |60−t| =
|60 − t|−3 . We choose to take µ(t) = (60 − t)−3 , without the absolute values, since we
expect solutions only for 0 ≤ t ≤ 60.
Multiplying the DE by this integrating factor, we have
0
1
1
3
2
x =
x0 +
x=
.
3
3
4
(60 − t)
(60 − t)
(60 − t)3
(60 − t)
R
1
2
1
Integrating, (60−
x
=
dt = (60−
+ C, so x = 60 − t + C (60 − t)3 .
3
t)
(60−t)3
t )2
Finally, we note that there is also an initial condition: x (0) = 0, since the tank starts with
no salt. Thus, 0 = x (0) = 60 − 0 + C (60 − 0)3 = 60 + 603 C, so C = 1/602 , and the total
amount of salt in the tank at time t is
x (t) = 60 − t −
1
(60 − t)3 .
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Solution (b): We find the maximum value of x (t) on [0, 60]. Computing x 0 (t),
x 0 ( t ) = −1 −
1
1
(3)(−1)(60 − t)2 = −1 +
(60 − t)2 .
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1200
At a local extremum, x 0 (t) = 0, since x 0 (t) exists everywhere
on this interval. Then 0 =
√
1
2
2
−1 + 1200 (60 − t) , so (60 − t) = 1200, and t = 60 − 20 3. Finally,
√
√
x (60 − 20 3) = 20 3 −
√
√
1
1 √
40 √
(20 3)3 = 20 3 − 20 3 =
3 ≈ 23.09 lbs.
3600
3
3
Since x (0) = x (60) = 0, this is the maximum amount of salt in the tank.