Download - Problems in Linear Programming 2
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
SolutionSee figure 1Draw the lines
4x1 + 6x2 = 360
3x1 = 180
5x2 = 200
z is maximum at a vertex of the feasible region, which isa polygon. That is at O,A,B,C, orD.
Extreme Point Coordinates zO (0, 0) 0A (60, 0) 900B (60, 20) 1,100C (30, 40) 850D (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
SolutionSee 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 isat O,A,B,C,D, orE.Extreme Point Coordinates z
O (0, 0) 0A (1, 0) 2B (3,1) 7C (4, 2) 10D (2,4) 8E (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
SolutionSee 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 isat A,B, orC.Extreme Point Coordinates z
A (0,1) 2B (0,2) 4C (2,3) 4
Max z is at the point B or C and hence at any pointbetween B and C, on line BC, and
max z = 4
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