BBMP1103 - Sept 2011 exam workshop - part 7

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BBMP1103 - Sept 2011 exam workshop - part 7

  1. 1. BBMP 1103 Mathematic Management Exam Preparation Workshop Sept 2011Part 7 - Application of Partial Differentiation Presented By: Dr Richard Ng 26 Nov 2011 2ptg – 4ptg
  2. 2. 7. Focus on Application of Partial Differentiation Question: 15 (May 2010)Prepared by Dr Richard Ng (2011) Page 2
  3. 3. Suggested Answers: i) f ( x, y) 25 2 x 3 y x 2 y2 xy fx 2 2x y 0 … (i) fy 3 2y x 0 … (ii) (i) x 2: 4 4x 2 y 0 … (iii) (iii) - (ii): 3x 7 0 7 x 3Prepared by Dr Richard Ng (2011) Page 3
  4. 4. Substitute into (i): 7 2 2 y 0 3 14 2 y 0 3 8 y 0 3 8 y 3 7 8 Hence, the critical point is , 3 3Prepared by Dr Richard Ng (2011) Page 4
  5. 5. ii) To determine maximum or minimum use M Test: fx 2 2x y fy 3 2y x f xx 2 f yy 2 f xy 1 M [( f xx )( f yy )] [( f xy )]2 [(2)(2)] [1]2 [4] [1] 3 Since M > 0 and fxx > 0, hence the critical point is minimumPrepared by Dr Richard Ng (2011) Page 5
  6. 6. Question: 16 (Sept 2009)Prepared by Dr Richard Ng (2011) Page 6
  7. 7. Suggested Answers: i) f ( x, y) 8x 2 8 y 2 8 xy 48x 5 fx 16x 8 y 48 0 … (i) fy 16 y 8x 0 … (ii) Equation (i) x 2: 32x 16 y 96 0 … (iii) (iii) – (ii) : 24x 96 0 x 4Prepared by Dr Richard Ng (2011) Page 7
  8. 8. Substitute x = -4 into (ii): 16y 8( 4) 0 16y 32 y 2 Hence, the critical point is = (-4, 2) ii) To determine maximum or minimum use M Test: fx 16x 8 y 48 fy 16 y 8 x f xx 16 f yy 16 f xy 8Prepared by Dr Richard Ng (2011) Page 8
  9. 9. M [( f xx )( f yy )] [( f xy )]2 [(16)(16)] [8]2 [256] [64] 192 Since M > 0 and fxx > 0, hence the critical point is minimum iii) When x = - 4, y = 2: f ( x, y) 8( 4) 2 8(2) 2 8( 4)(2) 48( 4) 5 128 32 64 192 5 101 Hence, the minimum value is = -101Prepared by Dr Richard Ng (2011) Page 9
  10. 10. End ofPart 7

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