Question

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

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)

200 x + 3y <120 -150 100 50 (0,40) (30,30) -400 -200 0 200 400 600 800 (40,0) --50 3x + y = 120 -- 100

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

x + y 10 10 - x + y<5 (2,7) 6 (4,6) 2x + 4y < 32 (-5,0) (10,0) 5 15 20 25

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.