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# week-7-class-2-solutions

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```Problem Set – Week 7 – Class 2
Solutions
ICGN104 Mathematics and Its Contemporary
Applications
1. The total-cost function for a hosiery mill is estimated to be
c = −10, 484.69 + 6.750q − 0.000328q 2 ,
where q is output in dozens of pairs and c is total cost in dollars. Find the marginal-cost
function and the average cost function and evaluate each when q = 2000.
Solution: The marginal-cost function is
dc
= 6.750 − 0.000328(2q) = 6.750 − 0.000656q.
dq
Thus, the marginal cost when q = 2000 is
󰀏
dc 󰀏󰀏
= 6.750 − 0.000656(2000) = 5.438.
dq 󰀏q=2000
This means increasing the output from 2000 to 2001 dozens of pairs will increase the
total cost by approximately \$5.438. The average cost function is
c=
c
10, 484.69
=−
+ 6.750 − 0.000328q.
q
q
Consequently, the average cost when q = 2000 is
c(2000) =
10, 484.69
+ 6.750 − 0.000328(2000) = 0.851655.
2000
2. The total-cost function for an electric light and power plant is estimated to be
c = 32.07 − 0.79q + 0.02142q 2 − 0.0001q 3 ,
20 ≤ q ≤ 90,
where q is the eight-hour total output (as a percentage of capacity) and c is the total fuel
cost in dollars. Find the marginal-cost function and evaluate it when q = 70. Interpret
the result.
Solution: The marginal-cost function is
dc
= −0.79 + 0.04284q − 0.0003q 2 .
dq
Thus, when q = 70, the marginal cost is
󰀏
dc 󰀏󰀏
= −0.79 + 0.04284(70) − 0.0003(70)2 = 0.7388.
dq 󰀏q=70
This means that increasing the eight-hour total output from q = 70 to q = 71 will
increase the total fuel cost by approximately 74 cents.
3. For the cost function
c = 0.3q 2 + 3.5q + 9,
how fast does c change with respect to q when q = 10? Determine the percentage rate
of change of c with respect to q when q = 10.
Solution: We want to determine the rate of change of c with respect to q. This is
given by c′ (q) = 0.6q + 3.5. When q = 10, we have c′ (10) = 0.6(10) + 3.5 = 9.5 and
c(10) = 74. Thus, the percentage rate of change of c with respect to q is given by
c′ (10)
9.5
&middot; 100% =
&middot; 100% ≈ 12.8%.
c(10)
74
4. For a certain manufacturer, the revenue obtained from the sale of q units of a product
is given by
r = 30q − 0.3q 2 .
(a) How fast does r change with respect to q?
(b) When q = 10, find the relative rate of change of r, and to the nearest percent, find
the percentage rate of change of r.
Solution:
(a) We want to find the rate of change of r with respect to q. This is given by
r′ (q) = 30 − 0.6q.
(b) When q = 10, we have r′ (10) = 30 − 6 = 24 and r(10) = 300 − 30 = 270. Hence,
the relative rate of change of r when q = 10 is given by
r′ (10)
24
4
=
=
≈ 0.09.
r(10)
270
45
Page 2
(c) The percentage rate of change of r when q = 10 is given by
r′ (10)
&middot; 100% ≈ (0.09)(100%) = 9%.
r(10)
5. A manufacturer of mountain bikes has found that when 20 bikes are produced per day,
the average cost is \$200 and the marginal cost is \$150. Based on that information,
approximate the total cost of producing 21 bikes per day.
Solution: The total cost, c, of producing 20 bikes is c = qc = 20(200) = \$4000. The
marginal cost, \$150, is the approximate cost of one additional bike (when 20 bikes are
produced). Thus the approximate cost of producing 21 bikes is \$4000 + \$150 = \$4150 .
6. A taco stand usually sells 225 tacos per day at \$2 each. A business student’s research
tells him that for every \$0.15 decrease in the price, the stand will sell 20 more tacos per
day. Let x be the number of \$0.15 reductions in price. Determine the revenue function,
dR
R(x), in terms of x and find
.
dx
Solution: The revenue function, R(x), for the taco stand is given by
R(x) = (2 − 0.15x)(225 + 20x).
By the product rule, we have
dR
d
d
= (2 − 0.15x) (225 + 20x) + (225 + 20x) (2 − 0.15x)
dx
dx
dx
= (2 − 0.15x)(20) + (225 + 20x)(−0.15)
= 40 − 3x − 33.75 − 3x = 6.25 − 6x.
7. If the demand equation for a manufacturer’s product is
p=
1000
,
q+5
where p is in dollars, find the marginal-revenue function and evaluate it when q = 45.
Interpret the result.
Solution: The revenue function, r, is
r = pq =
1000
1000q
&middot;q =
.
q+5
q+5
Page 3
Thus, the marginal-revenue function is given by
d
d
(1000q) − (1000q) (q + 5)
dr
dq
dq
=
2
dq
(q + 5)
(q + 5)(1000) − (1000q)(1)
5000
=
=
2
(q + 5)
(q + 5)2
(q + 5)
and
󰀏
dr 󰀏󰀏
5000
5000
=
=
= 2.
󰀏
2
dq q=45 (45 + 5)
2500
This means that selling one additional unit beyond 45 results in approximately \$2 more
in revenue.
8. A manufacturer has determined that, for his product, the daily average cost (in hundreds
of dollars) is given by
324
5 19
c= 󰁳
+ + .
2
q + 35 q 18
(a) As daily production increases, the average cost approaches a constant dollar amount.
What is this amount?
(b) Determine the manufacturer’s marginal cost when 17 units are produced per day.
(c) The manufacturer determines that if production (and sales) were increased to 18
units per day, revenue would increase by \$275. Should this move be made? Why?
Solution:
(a) We want to determine lim c. We have
q→∞
lim c = lim
q→∞
q→∞
󰀓
󰁳
324
q 2 + 35
+
5 19 󰀔
19
19
+
=0+0+
= .
q 18
18
18
(b) The total-cost function, c, is
c = cq = 󰁳
324q
q2
+ 35
+5+
19
q.
18
Thus, the marginal-cost function is given by
󰁳
1
q 2 + 35(324) − 324q( 12 )(q 2 + 35)− 2 (2q) 19
dc
=
+
dq
q 2 + 35
18
and
󰀏
dc 󰀏󰀏
= 3.
dq 󰀏q=17
(c) From part (b) the increase in cost of the additional unit is approximately \$300.
Since the corresponding revenue increases by \$275, the move should not be made.
Page 4
9. A manufacturer’s average-cost function, in dollars, is given by
c=
500
.
ln(q + 20)
Find the marginal cost (rounded to two decimal places) when q = 50.
Solution: We have c = cq =
500q
. Thus
ln(q + 20)
dc
= 500 &middot;
dq
and
[ln(q + 20)](1) − q
󰀓
1 󰀔
q + 20
[ln(q + 20)]2
50
󰀏
ln 70 −
dc 󰀏󰀏
70 ≈ \$97.90.
= 500 &middot;
dq 󰀏q=50
(ln 70)2
10. For a firm, the daily output on the tth day of a production run is given by
q = 500(1 − e−0.2t ).
Find the rate of change of output q with respect to t on the tenth day.
Solution: We have
Thus,
dq
= 500(−e−0.2t )(−0.2) = 100e−0.2t .
dt
󰀏
dq 󰀏󰀏
= 100e−2 .
dt 󰀏t=10
Page 5
```