BBMP 1103Mathematic Management

Exam Preparation Workshop Sept 2011Part 3 - Application of Differentiation

Presented By: Dr Richard Ng

26 Nov 20112ptg – 4ptg

3. Focus on Application of Differentiation

Question: 7 (January 2010)

Prepared by Dr Richard Ng (2011) Page 3

i) Average Cost =q

qq

q

CC

8009608.0 2

qq

8009608.0

ii) 18009608.0 qqC

080008.0 2 qdq

Cd

2

80008.0

q

Suggested Answers:

Prepared by Dr Richard Ng (2011) Page 4

08.0

8002 q

100002 q

100q

33

2

2 16001600

qq

dq

Cd

When q = 100,

02

2

dq

Cd=> minimum

Hence, q = 100 minimizes the average cost

Prepared by Dr Richard Ng (2011) Page 5

qqC

8009608.0

iii) When q = 100,

)100(

80096)100(08.0 C

8968

112

Hence, the minimum value of average cost = RM112

Prepared by Dr Richard Ng (2011) Page 6

Question: 8 (September 2008)

Prepared by Dr Richard Ng (2011) Page 7

a) Revenue R = p x q

= (48 – 3q)(q)

= 48q – 3q2

b) 2348 qqR

0648 qdq

dR

q648

8q

Suggested Answers:

Prepared by Dr Richard Ng (2011) Page 8

qdq

dR648

062

2

dq

Rd => maximum

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

Question: 9 (January 2011)

a) pqR

qqR )460(

2460 qq

b) )(4120

qq

C

xqC

q4120

c) CRP

qqqP 4120460 2

120564 2 qq

d) 120564 2 qqP

0568 qdq

dP

7q

82

2

dq

Pd

Since 082

2

dq

PdHence the profit is maximum

When q = 7, 32)7(460 p

Hence, the price that maximizes the profit is RM32

End ofPart 3