Prepared by Dr. Richard Ng

1. BBMP 1103
Mathematic Management
Exam Preparation Workshop Sept 2011
Part 3 - Application of Differentiation
Presented By: Dr Richard Ng
26 Nov 2011
2ptg – 4ptg

2. 3. Focus on Application of Differentiation
Question: 7 (January 2010)

3. Suggested Answers:
C 0.08q 2 96q 800
i) Average Cost = C
q q
800
0.08q 96
q
1
ii) C 0.08q 96 800q
dC 2
0.08 800q 0
dq
800
0.08
q2
Prepared by Dr Richard Ng (2011) Page 3

4. 2 800
q
0.08
q2 10000
q 100
d 2C 3 1600
1600q
dq 2 q3
When q = 100,
d 2C
2
0 => minimum
dq
Hence, q = 100 minimizes the average cost
Prepared by Dr Richard Ng (2011) Page 4

5. iii) When q = 100,
800
C 0.08q 96
q
800
C 0.08(100) 96
(100)
8 96 8
112
Hence, the minimum value of average cost = RM112
Prepared by Dr Richard Ng (2011) Page 5

6. Question: 8 (September 2008)
Prepared by Dr Richard Ng (2011) Page 6

7. Suggested Answers:
a) Revenue R = p x q
= (48 – 3q)(q)
= 48q – 3q2
b) R 48q 3q 2
dR
48 6q 0
dq
48 6q
q 8
Prepared by Dr Richard Ng (2011) Page 7

8. dR
48 6q
dq
d 2R
6 0 => maximum
dq 2
Hence, q = 8 maximizes the revenue
c) When q = 8,
R = 48(8) – 3(8)2
R = 384 – 192 = 192
Hence, the maximum value of revenue = RM192
Prepared by Dr Richard Ng (2011) Page 8

9. Question: 9 (January 2011)

10. a) R pq
R (60 4q)q
60q 4q 2
120
b) C Cxq 4 (q)
q
120 4q
c) P R C
P 60q 4q 2 120 4q
4q 2 56q 120

11. d) P 4q 2 56q 120
dP
8q 56 0
dq
q 7
d 2P
2
8
dq
d 2P
Since 2
8 0 Hence the profit is maximum
dq
When q = 7, p 60 4(7) 32
Hence, the price that maximizes the profit is RM32

12. End of
Part 3