BBMP 1103
   Mathematic Management
Exam Preparation Workshop Sept 2011
     Part 8 - Lagrange Multiplier

       Presented By: Dr Richard Ng

              26 Nov 2011
                2ptg – 4ptg
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
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)
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
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)
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)
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)
Question: 18 (January 2010)
Question: 19 (September 2006)




Prepared by Dr Richard Ng (2011)   Page 9
End of
                                     Exam
                                   Workshop




Prepared by Dr Richard Ng (2011)              Page 10

BBMP1103 - Sept 2011 exam workshop - part 8

  • 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 onLagrange 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 (January2011) 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.
  • 9.
    Question: 19 (September2006) Prepared by Dr Richard Ng (2011) Page 9
  • 10.
    End of Exam Workshop Prepared by Dr Richard Ng (2011) Page 10