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# BBMP1103 - Sept 2011 exam workshop - part 8

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Prepared by Dr. Richard Ng

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### BBMP1103 - Sept 2011 exam workshop - part 8

1. 1. BBMP 1103 Mathematic ManagementExam Preparation Workshop Sept 2011 Part 8 - Lagrange Multiplier Presented By: Dr Richard Ng 26 Nov 2011 2ptg – 4ptg
2. 2. 8. Focus on Lagrange Multiplier Find the minimum value for f ( x, y) 5x 2 6 y 2 xy over a constraint of x + 2y = 24 Answers: Step: 1 – Express constraint in the form of g(x,y) = 0 x + 2y = 24 x + 2y – 24 = 0 g(x,y) = x + 2y – 24Prepared by Dr Richard Ng (2011) Page 2
3. 3. Step: 2 – Form the Lagrange function f ( x, y, ) F ( x, y, ) f ( x, y) g ( x, y) F ( x, y, ) (5x 2 6 y 2 xy) [ x 2 y 24] F ( x, y, ) 5x 2 6 y 2 xy x 2 y 24Step: 3 – Find Fx , Fy , F and equate to zero Fx 10x y 0 … (i) Fy 12y x 2 0 … (ii) F x 2y 24 0 … (iii)
4. 4. Step: 4 – Solve the 3 equations (i) x 2 => 20x 2 y 2 0 … (iv) (iv) - (ii) => 21x 14 y 0 21x 14 y 2 x y … (v) 3 Substitute (v) into (iii): 2 y 2 y 24 0 3
5. 5. 2 y 2 y 24 0 3 8 y 24 3 y 9Substitute into (v): 2 x (9) 3 x 6Hence, the minimum value is = (6, 9)
6. 6. Question: 17 (January 2011)Suggested Answers: x + 2y = 20 x + 2y – 20 = 0 Hence, g(x,y) = x + 2y – 20 F f ( x, y) g ( x, y) F [2 x 2 8 y 2 xy] [ x 2 y 20] Fx 4x y 0 … (i)
7. 7. Fy 16y x 2 0 … (ii) F x 2y 20 0 … (iii)(i) X 2 => 8 x 2 y 2 0 … (iv)(iv) – (ii) => 9 x 18y 0 x 2y … (v)Substitute (v) into (iii): (2 y) 2 y 20 0 4y 20 y 5 x 2y 2(5) 10Hence, the minimum value is => (10, 5)
8. 8. Question: 18 (January 2010)
9. 9. Question: 19 (September 2006)Prepared by Dr Richard Ng (2011) Page 9
10. 10. End of Exam WorkshopPrepared by Dr Richard Ng (2011) Page 10