BBMP1103 - Sept 2011 exam workshop - part 8
-
Upload
richard-ng -
Category
Education
-
view
4.922 -
download
4
description
Transcript of BBMP1103 - Sept 2011 exam workshop - part 8
![Page 1: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/1.jpg)
BBMP 1103Mathematic Management
Exam Preparation Workshop Sept 2011Part 8 - Lagrange Multiplier
Presented By: Dr Richard Ng
26 Nov 20112ptg – 4ptg
![Page 2: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/2.jpg)
Prepared by Dr Richard Ng (2011) Page 2
Answers:
Find the minimum value for xyyxyxf 22 65),(
over a constraint of x + 2y = 24
Step: 1 – Express constraint in the form of g(x,y) = 0
x + 2y = 24
x + 2y – 24 = 0
g(x,y) = x + 2y – 24
8. Focus on Lagrange Multiplier
![Page 3: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/3.jpg)
Step: 2 – Form the Lagrange function ),,( yxf
),(),(),,( yxgyxfyxF
]242[)65(),,( 22 yxxyyxyxF
Step: 3 – Find FFF yx ,, and equate to zero
24265),,( 22 yxxyyxyxF
010 yxFx … (i)
0212 xyFy … (ii)
0242 yxF … (iii)
![Page 4: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/4.jpg)
Step: 4 – Solve the 3 equations
(i) x 2 => 02220 yx … (iv)
(iv) - (ii) => 01421 yx
yx 1421
yx3
2 … (v)
Substitute (v) into (iii):
02423
2
yy
![Page 5: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/5.jpg)
02423
2
yy
243
8y
9y
Substitute into (v):
)9(3
2x
6x
Hence, the minimum value is = (6, 9)
![Page 6: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/6.jpg)
x + 2y = 20
x + 2y – 20 = 0
Hence, g(x,y) = x + 2y – 20
),(),( yxgyxfF
]202[]82[ 22 yxxyyxF
04 yxFx … (i)
Question: 17 (January 2011)
Suggested Answers:
![Page 7: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/7.jpg)
0216 xyFy … (ii)
0202 yxF … (iii)
(i) X 2 => 0228 yx … (iv)
(iv) – (ii) => 0189 yx
yx 2 … (v)
Substitute (v) into (iii):
0202)2( yy
204 y5y
10)5(22 yx
Hence, the minimum value is => (10, 5)
![Page 8: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/8.jpg)
Question: 18 (January 2010)
![Page 9: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/9.jpg)
Question: 19 (September 2006)
Prepared by Dr Richard Ng (2011) Page 9
![Page 10: BBMP1103 - Sept 2011 exam workshop - part 8](https://reader035.fdocuments.in/reader035/viewer/2022081203/54bc61b74a795943628b45c9/html5/thumbnails/10.jpg)
End ofExam
Workshop
Prepared by Dr Richard Ng (2011) Page 10