Chapter 19 – Linear Programming
3. Maximize: 40H + 30W
Subject to:
Fabrication 4H + 2W 600 hours
Assembly 2H + 6W 480 hours
a. Optimum:
H = 132
W = 36
Z = $6,360
The work for this solution is shown below:
Fabrication constraint:
4H + 2W 600
Replace the inequality sign with an equal sign:
4H + 2W = 600
Set H = 0 and solve for W:
4(0) + 2W = 600
Assembly constraint:
2H + 6W 480
W = 80
One point on the line is (0, 80).
Set W = 0 and solve for H:
2H + 6(0) = 480
2H = 480
Chapter 19 – Linear Programming
Objective function:
Let 40H + 30W = 4,800.
Set H = 0 and solve for W:
40(0) + 30W = 4,800
40H = 4,800
H = 120
A second point on the line is (120, 0).
As we slide the profit line away from the origin, we reach the optimum point shown in the
graph below:
Optimum
Profit
300
250
200
150
100
50
0
100 150 200 250
Assembly
W
H
Fabrication
Chapter 19 – Linear Programming
The optimum occurs at the intersection of the Fabrication and Assembly constraints.
result:
12H + 6W = 1,800
–(2H + 6W = 480)
10H = 1,320
H = 132
Step 2:
Substitute H = 132 in either constraint:
W = 36
Step 3:
Substitute the values of H and W in the objective function:
Optimum:
H = 132
Chapter 19 – Linear Programming
19–24
4. Peanuts cost $.60/lb. Deluxe revenue is $2.90/lb.
Raisins cost $1.50/lb. Standard revenue is $2.5 /lb.
Deluxe mix has 1/3 lb. peanuts & 2/3 lb. raisins.
Hence, the Deluxe mix cost is:
Standard mix has ½ lb. peanuts & ½ lb. raisins.
Hence, the Standard mix cost is:
a. Optimum:
b. Z = $243
The work for these solutions is shown below:
Chapter 19 – Linear Programming
Raisins constraint:
Replace the inequality sign with an equal sign:
Peanuts constraint:
Replace the inequality sign with an equal sign:
Set D = 0 and solve for S:
One point on the line is (0, 120).
Set S = 0 and solve for D:
Standard constraint:
1D ≤ 110
Replace the inequality sign with an equal sign:
1D = 110
Deluxe constraint:
Chapter 19 – Linear Programming
Objective function:
Let 1.70D + 1.50S = 127.5.
Set D = 0 and solve for S:
1.70D + 1.50(0) = 127.5
1.70D = 127.5
D = 75
As we slide the profit line away from the origin, we reach the optimum point shown in the
graph below:
Peanuts
Profit
S
D = 110
S = 110
Raisins
220
200
180
160
140
120
Chapter 19 – Linear Programming
Simultaneous solution:
Step 1:
Subtract the Peanuts constraint from the Raisins constraint:
________________
D = 90
Step 2:
S = 60
Step 3:
Substitute the values of D & S in the objective function:
Chapter 19 – Linear Programming
19–28
5. Maximize: 1.50A + 1.20G
Subject to:
Sugar: 1.5A + 2.0G 1,200 cups
Flour: 3.0A + 3.0G 2,100 cups
Time: 6.0A + 3.0G 3,600 min.
a. Optimum:
A = 500 apple pies
G = 200 grape pies
Revenue = $990
The work for these solutions is shown below:
Sugar constraint:
1.5A + 2.0G 1,200
Replace the inequality sign with an equal sign:
1.5A + 2.0G = 1,200
Set A = 0 and solve for G:
1.5(0) + 2.0G = 1,200
2.0G = 1,200
G = 600
One point on the line is (0, 600).
Flour constraint:
3.0A + 3.0G 2,100
Replace the inequality sign with an equal sign:
3.0A + 3.0G = 2,100
Set A = 0 and solve for G:
3.0(0) + 3.0G = 2,100
3.0G = 2,100
Chapter 19 – Linear Programming
Time constraint:
6.0A + 3.0G 3,600
Replace the inequality sign with an equal sign:
6.0A + 3.0G = 3,600
Set A = 0 and solve for G:
Objective function:
Let 1.50A + 1.20G = 600
Set A = 0 and solve for G:
1.50(0) + 1.20G = 600
1.20G = 600
G = 500
One point on the line is (0, 500).
G
Time
1200
1000
800
600
Flour
Chapter 19 – Linear Programming
19–30
The optimum point occurs at the intersection of the Flour and Time constraints.
Step 1:
Subtract the Time constraint from the Flour Constraint:
3.0A + 3.0G = 2,100
Step 2:
Substitute A = 500 in either constraint:
3.0A + 3.0G = 2,100
3.0(500) + 3.0G = 2,100
Step 3:
Substitute the values of A & G in the objective function:
b. Amount of sugar, flour, and time that will be unused:
The Time and Flour constraints are binding; therefore, there will be no unused time or
flour.
To determine the unused amount of sugar, we must plug in the values of A & G in the
Sugar constraint:
1.5A + 2.0G 1,200