Solve by Linear Programming.
(Be sure to show the graph of the feasible region, the appropriate vertices, optimal value,
AND SHOW ALL WORK!.)
Exercise 1
LP 1. Maximize:
C = x – y
Constraints:
x ≥ 0, and y ≥ 0
x + 3y ≤ 120
3x + y ≤ 120
Exercise 2
LP 2. Maximize:
C = 3x + 4y
Constraints:
x + y ≤ 10
– x + y ≤ 5
2x + 4y ≤ 32
Answer:
1) Maximize = x-y
Constraints, x ≥ 0, and y ≥ 0
x + 3y ≤ 120
3x + y ≤ 120
Finding coordiantes of the coordiantes,
x + 3y ≤ 120
x | 0 | 120 |
y | 40 | 0 |
3x + y ≤ 120
x | 0 | 40 |
y | 120 | 0 |
The common shaded region is the feasible region. The vertices of the feasible region are the possible optimal solution. The vertices are : (0,40) , (40,0) and (30,30)
Maximize = x-y
(0,40) = 0-40 = -40
(40,0) = 40-0 = 40
(30,30) = 30-30 = 0
Since the (40,0) gives the maximum value, therefore, it is the optimal solution.
2)
Maximize = 3x + 4y
Constraints, x + y ≤ 10
– x + y ≤ 5
2x + 4y ≤ 32
Finding coordiantes of the coordiantes,
x + y ≤ 10
x | 0 | 10 |
y | 10 | 0 |
– x + y ≤ 5
x | 0 | -5 |
y | 5 | 0 |
2x + 4y ≤ 32
x | 0 | 16 |
y | 8 | 0 |
The vertices of the feasible region are the possible optimal solution. The vertices are : (-5,0) , (10,0) , (4,6) and (2,7)
Maximize = 3x + 4y
(-5,0) = 3(-5)+4(0) = -15
(10,0) = 3(10) + 4(0) = 30
(4,6) = 3(4) + 4(6)=36
(2,7) = 3(2)+4(7)=34
Since the (4,6) gives the maximum value, therefore, it is the optimal solution.
Solve by Linear Programming. (Be sure to show the graph of the feasible region, the appropriate...
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