Prepared by Dr. Richard Ng

1. BBMP 1103
Mathematic Management
Exam Preparation Workshop Sept 2011
Part 8 - Lagrange Multiplier
Presented By: Dr Richard Ng
26 Nov 2011
2ptg – 4ptg

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 – 24
Prepared by Dr Richard Ng (2011) Page 2

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 24
Step: 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. 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. 2
y 2 y 24 0
3
8
y 24
3
y 9
Substitute into (v):
2
x (9)
3
x 6
Hence, the minimum value is = (6, 9)

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. 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) 10
Hence, the minimum value is => (10, 5)

8. Question: 18 (January 2010)

9. Question: 19 (September 2006)
Prepared by Dr Richard Ng (2011) Page 9

10. End of
Exam
Workshop
Prepared by Dr Richard Ng (2011) Page 10