BBMP 1103   Mathematic ManagementExam Preparation Workshop Sept 2011  Part 5 - Application of Integration        Presented...
5. Focus on Application of Integration     Question: 12 (May 2010)Prepared by Dr Richard Ng (2011)              Page 2
Suggested Answers:           i) Given                C 30x                Hence, C              30x.dx                    ...
ii) When x = 100:                            C 15(100) 2 15000                                   150000 15000             ...
Question: 13 (Jan 2010)Prepared by Dr Richard Ng (2011)   Page 5
Suggested Answers:                                   80000           i) Given C                                     x2    ...
When x = 0, c = 27000                                        80000                Hence, C                      27000     ...
When x = 400                                        80000                                   C          27000              ...
End ofPart 5
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BBMP1103 - Sept 2011 exam workshop - part 5

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BBMP1103 - Sept 2011 exam workshop - part 5

  1. 1. BBMP 1103 Mathematic ManagementExam Preparation Workshop Sept 2011 Part 5 - Application of Integration Presented By: Dr Richard Ng 26 Nov 2011 2ptg – 4ptg
  2. 2. 5. Focus on Application of Integration Question: 12 (May 2010)Prepared by Dr Richard Ng (2011) Page 2
  3. 3. Suggested Answers: i) Given C 30x Hence, C 30x.dx x2 30 c 2 15x 2 c When x = 0, c = 15000 Hence, C 15x 2 15000Prepared by Dr Richard Ng (2011) Page 3
  4. 4. ii) When x = 100: C 15(100) 2 15000 150000 15000 165000 When x = 150: C 15(150) 2 15000 15(22500) 15000 352500 Hence, increase in cost = 352500 - 165000 = RM187,500Prepared by Dr Richard Ng (2011) Page 4
  5. 5. Question: 13 (Jan 2010)Prepared by Dr Richard Ng (2011) Page 5
  6. 6. Suggested Answers: 80000 i) Given C x2 80000 Hence, C 2 .dx x 80000x 2 .dx 1 80000x c 1 80000 c xPrepared by Dr Richard Ng (2011) Page 6
  7. 7. When x = 0, c = 27000 80000 Hence, C 27000 x ii) When x = 200 80000 C 27000 (200) 400 27000 26600Prepared by Dr Richard Ng (2011) Page 7
  8. 8. When x = 400 80000 C 27000 (400) 200 27000 26800 Hence, increase in cost = 26800 - 26600 = RM200Prepared by Dr Richard Ng (2011) Page 8
  9. 9. End ofPart 5
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