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1.0Linear Programming Solved Examples
Example 1:
A furniture dealer specializes in selling only two products: tables and chairs. He has Rs. 5000 available to invest and can store up to 60 pieces in total. The cost to purchase a table is Rs. 250, while a chair costs Rs. 50. The dealer can make a profit of Rs. 50 on each table and Rs. 15 on each chair. Assuming that he can sell all the items he purchases, how should he allocate his investment to maximize his profit?
Solution:
This problem can be formulated as below.
Let x and y be the required numbers of tables and chairs respectively. Then, clearly, we have
x ≥ 0,
y ≥ 0;
x + y ≤ 60;
250x + 50y ≤ 5000. i.e., 5x+ y ≤ 100
Let P be the profit function. Then, P = 50 + 15y.
Now, we have to maximize P
Now, x+y=60⇒60x+60y=1.
This line meets the axes at (60, 0) and (0, 60). Plot these points and join them to get the line x + y = 60
Also, 5x + y = 100⇒20x+100y=1 .
This line meets the axes at (20, 0) and (0, 100). Plot these points and
join them to get the line 5x + y = 100.
Also, the line x = 0 is the y-axis and the line y = 0 is the x-axis.
These four straight lines enclose the quadrilateral OABC.
The coordinates of the points O, A, B, C are (0, 0), (20, 0), (10, 50) and (0, 60) respectively.
At these points, the corresponding values of P = 50x + 50 +15y are 0, 1000, 1250 and 900 respectively. Clearly, it is maximum at B (10, 50).
In other words, we can say that
Maximize. P = 50x + 15y
X + y ≤ 6 0
250x + 50y ≤ 5000
x ≥ 0, y ≥ 0.
Points | Corner points | Corresponding Value of F |
O | (0, 0) | 0 |
A | (20,0) | 1000 |
B | (10,50) | 1250 → Maximize |
C | (0,60) | 900 |
So, for a maximum profit, the dealer should purchase 10 tables and 50 chairs
Example 2:
A young man rides his motorcycle at two different speeds: 25 km per hour, costing him Rs. 2 per kilometer for petrol, and 40 km per hour, costing Rs. 5 per kilometer. With Rs. 100 available for petrol, he wants to determine the maximum distance he can cover within one hour. Frame this scenario as a linear programming problem and then solve it.
Solution:
Suppose that the young man rides x km at 25 km per hour and y km at 40 km per hour. Then, we have to maximize P = x + y.
Clearly:
x ≥ 0,
y ≥ 0,
2x + 5y ≤ 100.
Since the available time is at most one hour, we have
25x+40y≤1 or 8x + 5y ≤ 200
Now, we solve the system of the inequations.
2x + 5y = 100 ⇒50x+20y=1 .
This line meets the axes at (50, 0) and (0, 20). Plot these points and join them to get the line 2x + 5y = 100.
In other words,
Maximize Z = x + y
8x + 5y ≤ 200
2x + 5y ≤ 100
x ≥ 0, y ≥ 0
Points | Corner points | Corresponding Value of Z |
O | (0, 0) | 0 |
A | (25,0) | 25 |
B | \left(\frac{50}{3}, \frac{40}{3}\right) | 30 ← Maximize |
C | (0,20) | 20 |
So, Z = x + y is maximum when x=350 and y=340 . Thus, the young man can cover the maximum distance of 30 km, if he rides 350 km at 25 km/h and 340 km at 40 km/h
Example 3:
Suppose every gram of wheat provides 01. g of proteins and 0 25. g of carbohydrates, and the corresponding values for rice are 0 05. g and 0 5. g respectively. Wheat costs RS. 5 and rice RS. 20 per kilogram. The minimum daily requirements of proteins and carbohydrates for an average man are 50 g and 200 g respectively. In what quantities should wheat and rice be combined in the daily diet to provide the minimum daily requirements of proteins and carbohydrates at minimum cost, assuming that both wheat and rice are to be taken in the diet?
Solution:
Let x g of wheat and y g of rice be mixed to fulfill the requirements. Then, we have to minimize the cost function.
Z=10005x+100020y, i.e., Z=200x+50y …(i)
x g of wheat and y g of rice must give at least 50 g of proteins.
So, we must have
0.1 x + 0.05 y ≥ 50 or 2x + y ≥ 1000.
Similarly, x g of wheat and y g of rice must give at least 200 g of carbohydrates.
So, we must have
0.25x +0.5y ≥ 200 or x + 2y ≥ 800.
Thus, we have to minimize
Z=200x+50y , subject to the constraints
x > 0,
y > 0,
2x + y ≥ 1000 and
x + 2y ≥ 800.
2x + y = 1000 ⇒ 500x+1000y=1
Minimize
Z=10005x+100020y
i.e. Z=200x+50y
2x +y ≥ 1000
x +2y ≥ 800
Points | Corner points | Corresponding Value of Z |
C | (800,0) | 4 |
E | (400, 200) | 6 |
B | (1000, 0) | 4 |
Clearly, at B we have x = 0, and at C we have y = 0. Also, the value of Z at E (400, 200) = 6. So, we must have 400 g of wheat and 200 g of rice
Example 4:
A chemical industry manufactures two compounds, A and B. The table below shows the amount of ingredients C and D (in units per kg) required for compounds A and B, along with the minimum requirements for C and D, and the costs per kg of compounds A and B.
Compound (in units) | Minimum requirement (in units) | ||
Ingredient C (per kg) | 1 | 2 | 80 |
Ingredient D (Per kg) | 3 | 1 | 75 |
Cost per Kg (in (Rs) | 4 | 6 | |
Find the quantities of A and B which would minimize the cost.
Solution:
Let x kg of A and y kg of B be produced. Then,
x ≥ 0, y ≥ 0, x + 2y ≥ 80 and 3x +y ≥ 75.
Then the cost function is given by Z = 4x + 6y.
Minimize Z = 4x + 6y.
Subject to the constraints.
X +2y ≥ 80
3x +y ≥ 75
x ≥ 0. y ≥ 0.
Points | Corner points | Corresponding Value of Z |
P | (0,75) | 450 |
Q | (14,33) | 254 ← Minimum |
R | (80, 0) | 320 |
Thus Z is minimum at Q (14,33)
Hence, for a minimum cost, 14 kg of a and 33 kg of B must be taken.
2.0Linear Programming Practice Problems
1. A man has Rs. 1500 to spend on purchasing rice and wheat. A bag of rice costs Rs. 180, and a bag of wheat costs Rs. 120. With a storage capacity of only 10 bags, he earns a profit of Rs. 11 per bag of rice and Rs. 8 per bag of wheat. How many bags of each should he buy to maximize his profit?
2. A manufacturer produces nuts and bolts for industrial machinery. To produce a packet of nuts, it requires 1 hour on machine A and 3 hours on machine B, while producing a packet of bolts requires 3 hours on machine A and 1 hour on machine B. The manufacturer earns a profit of Rs. 17.50 per packet of nuts and Rs. 7 per packet of bolts. If the machines can operate for a maximum of 12 hours each day, how many packets of each should be produced daily to maximize profit? Additionally, calculate the maximum profit achievable.
3. A dealer wishes to purchase a number of fans and sewing machines. He has only ` Rs.5760 to invest and has space for at most 20 items. A fan costs him Rs. 360 and a sewing machine, Rs. 240. He expects to gain Rs.22 on a fan and Rs.18 on a sewing machine. Assuming that he can sell all the items he can buy, how should he invest the money in order to maximize the profit?
Also Read : Linear Inequalities Questions