The second group of a series of problems in LP.

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
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