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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