BBMP1103 - Sept 2011 exam workshop - part 3

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BBMP1103 - Sept 2011 exam workshop - part 3

  1. 1. BBMP 1103 Mathematic ManagementExam Preparation Workshop Sept 2011 Part 3 - Application of Differentiation Presented By: Dr Richard Ng 26 Nov 2011 2ptg – 4ptg
  2. 2. 3. Focus on Application of DifferentiationQuestion: 7 (January 2010)
  3. 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 q2Prepared by Dr Richard Ng (2011) Page 3
  4. 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 costPrepared by Dr Richard Ng (2011) Page 4
  5. 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 = RM112Prepared by Dr Richard Ng (2011) Page 5
  6. 6. Question: 8 (September 2008)Prepared by Dr Richard Ng (2011) Page 6
  7. 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 8Prepared by Dr Richard Ng (2011) Page 7
  8. 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 = RM192Prepared by Dr Richard Ng (2011) Page 8
  9. 9. Question: 9 (January 2011)
  10. 10. a) R pq R (60 4q)q 60q 4q 2 120b) C Cxq 4 (q) q 120 4qc) P R C P 60q 4q 2 120 4q 4q 2 56q 120
  11. 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. 12. End ofPart 3

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