BBMP 1103Mathematic Management

Exam Preparation Workshop Sept 2011Part 8 - Lagrange Multiplier

Presented By: Dr Richard Ng

26 Nov 20112ptg – 4ptg

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

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)

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

02423

2

yy

243

8y

9y

Substitute into (v):

)9(3

2x

6x

Hence, the minimum value is = (6, 9)

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:

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)

Question: 18 (January 2010)

Question: 19 (September 2006)

Prepared by Dr Richard Ng (2011) Page 9

End ofExam

Workshop

Prepared by Dr Richard Ng (2011) Page 10