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PROBLEMS IN LINEAR PROGRAMMING 2


                   SEBASTIAN VATTAMATTAM




                 1. Graphical Solution
Problem 1.1. Solve graphically:
Maximize
                      z = 15x1 + 10x2
subject to the constraints
  (1) 4x1 + 6x2 ≤ 360
  (2) 3x1 + 0x2 ≤ 180
  (3) 0x1 + 5x2 ≤ 200
and x1 , x2 ≥ 0
  Solution
See figure 1
Draw the lines
                    4x1 + 6x2 = 360
                          3x1 = 180
                          5x2 = 200
  z is maximum at a vertex of the feasible region, which is
a polygon. That is at O, A, B, C, orD.
  Extreme Point Coordinates       z
        O             (0, 0)      0
        A            (60, 0)     900
        B           (60, 20)    1,100
        C           (30, 40)     850
        D            (0, 40)     400

 Max z = 1, 100 at the point (60, 20).
                             1
2                         SEBASTIAN VATTAMATTAM




                          Figure 1. Problem 1.1



Problem 1.2. Solve graphically:
Maximize
                             z = 2x1 + x2
subject to the constraints
    (1)   x1 + 2x2 ≤ 10
    (2)   x1 + x2 ≤ 6
    (3)   x1 − x2 ≤ 2
    (4)   x1 − 2x2 ≤ 1
and x1 , x2 ≥ 0

  Solution
See figure 2
LINEAR PROGRAMMING                   3




                    Figure 2. Problem 1.2


 Draw the lines
                     x1 + 2x2    =   10
                      x1 + x2    =   6
                      x1 − x2    =   2
                     x1 − 2x2    =   1
  z is maximum at a vertex of the feasible region. That is
at O, A, B, C, D, orE.
 Extreme Point Coordinates z
        O            (0, 0)    0
        A            (1, 0)    2
        B            (3,1)     7
        C            (4, 2)   10
        D            (2,4)     8
        E            (0, 5)    5
  Max z = 10 at the point (4, 2).
4                    SEBASTIAN VATTAMATTAM


Problem 1.3. Solve graphically:
Maximize
                          z = −x1 + 2x2
subject to the constraints
    (1) x1 − x2 ≤ −1
    (2) −0.5x1 + x2 ≤ 2
and x1 , x2 ≥ 0




                      Figure 3. Problem 1.3


  Solution
See figure 3

    Draw the lines
                        x1 − 2x2 = −1
                     −0.5x1 + x2 = 2
LINEAR PROGRAMMING                     5


  z is maximum at a vertex of the feasible region. That is
at A, B, orC.
 Extreme Point Coordinates z
        A          (0,1)     2
        B          (0,2)     4
        C          (2,3)     4
  Max z is at the point B or C and hence at any point
between B and C, on line BC, and
                       max z = 4

  For online classes in Mathematics at any level, please
                         contact
                 vattamattam@gmail.com
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Problems in Linear Programming 2

  • 1. PROBLEMS IN LINEAR PROGRAMMING 2 SEBASTIAN VATTAMATTAM 1. Graphical Solution Problem 1.1. Solve graphically: Maximize z = 15x1 + 10x2 subject to the constraints (1) 4x1 + 6x2 ≤ 360 (2) 3x1 + 0x2 ≤ 180 (3) 0x1 + 5x2 ≤ 200 and x1 , x2 ≥ 0 Solution See figure 1 Draw the lines 4x1 + 6x2 = 360 3x1 = 180 5x2 = 200 z is maximum at a vertex of the feasible region, which is a polygon. That is at O, A, B, C, orD. Extreme Point Coordinates z O (0, 0) 0 A (60, 0) 900 B (60, 20) 1,100 C (30, 40) 850 D (0, 40) 400 Max z = 1, 100 at the point (60, 20). 1
  • 2. 2 SEBASTIAN VATTAMATTAM Figure 1. Problem 1.1 Problem 1.2. Solve graphically: Maximize z = 2x1 + x2 subject to the constraints (1) x1 + 2x2 ≤ 10 (2) x1 + x2 ≤ 6 (3) x1 − x2 ≤ 2 (4) x1 − 2x2 ≤ 1 and x1 , x2 ≥ 0 Solution See figure 2
  • 3. LINEAR PROGRAMMING 3 Figure 2. Problem 1.2 Draw the lines x1 + 2x2 = 10 x1 + x2 = 6 x1 − x2 = 2 x1 − 2x2 = 1 z is maximum at a vertex of the feasible region. That is at O, A, B, C, D, orE. Extreme Point Coordinates z O (0, 0) 0 A (1, 0) 2 B (3,1) 7 C (4, 2) 10 D (2,4) 8 E (0, 5) 5 Max z = 10 at the point (4, 2).
  • 4. 4 SEBASTIAN VATTAMATTAM Problem 1.3. Solve graphically: Maximize z = −x1 + 2x2 subject to the constraints (1) x1 − x2 ≤ −1 (2) −0.5x1 + x2 ≤ 2 and x1 , x2 ≥ 0 Figure 3. Problem 1.3 Solution See figure 3 Draw the lines x1 − 2x2 = −1 −0.5x1 + x2 = 2
  • 5. LINEAR PROGRAMMING 5 z is maximum at a vertex of the feasible region. That is at A, B, orC. Extreme Point Coordinates z A (0,1) 2 B (0,2) 4 C (2,3) 4 Max z is at the point B or C and hence at any point between B and C, on line BC, and max z = 4 For online classes in Mathematics at any level, please contact [email protected]