Class 12 RD Sharma Solutions – Chapter 30 Linear Programming – Exercise 30.5
Question 1. Two godowns A and B have gain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs) | ||
---|---|---|
From/ To | A Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the Demo Class for First Step to Coding Course, specifically designed for students of class 8 to 12. The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future. | B |
D | 6 | 4 |
E | 3 | 2 |
F | 2.50 | 3 |
How should the supplies be transported in order that the transportation cost is minimum? what is the minimum cost?
Solution:
Let us assume godown A supply x and y quintals of grain to the shops D and E.
Afterwards, (100 – x – y) will be supplied to shop F.
The requirement at shop D is 60 quintals
Since, x quintals are transported from godown A.
Therefore, the remaining (60 − x) quintals will be transported from godown B.
Similarly,
(50 − y) quintals and 40 − (100 − x − y) i.e. (x + y − 60) quintals will be transported from godown B to shop E and F respectively.
The diagrammatic representation of the given problem:
x ≥ 0 , y ≥ 0 and 100 – x – y ≥ 0
⇒ x ≥ 0 , y ≥ 0 , and x + y ≤ 100
60 – x ≥ 0 , 50 – y ≥ 0 , and x + y – 60 ≥ 0
⇒ x ≤ 60 , y ≤ 50 , and x + y ≥ 60
Total transportation cost Z is given by,
Z = 6x + 3y + 2.5(100 – x – y) + 4(60 – x) + 2(50 – y) + 3( x + y – 60)
= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x + 100 – 2y + 3x + 3y – 180
= 2.5x + 1.5y + 410
Thus, the required mathematical formulation of linear programming is:
Minimize Z = 2.5x + 1.5y + 410
subject to the constraints,
x + y ≤ 100
x ≤ 60
y ≤ 50
x + y ≥ 60
x, y ≥ 0
The feasible region obtained by the system of constraints is:
The corner points are A(60, 0), B(60, 40), C(50, 50), and D(10, 50).
The values of Z at these corner points are given below.
Corner point Z = 2.5x + 1.5y + 410 A (60, 0) 560 B (60, 40) 620 C (50, 50) 610 D (10, 50) 510 -> minimum Therefore,
The minimum value of Z is 510 at D (10, 50).
Hence, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively.
From B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.
The minimum cost is Rs 510.
Question 2. A medical company has factories at two places, A and B. From these places, supply is made to each of its three agencies situated at P, Q and R. The monthly requirements of the agencies are respectively 40, 40 and 50 packets of the medicines, while the production capacity of the factories, A and B are 60 and 70 packets respectively. The transportation cost per packet from the factories to the agencies are given below:
Transportation cost per packet (in Rs) | ||
To\From | A | B |
P | 5 | 4 |
Q | 4 | 2 |
R | 3 | 5 |
How many packets from each factory be transported to each agency so that the cost of transportation is minimum? Also, find the minimum cost?
Solution:
The diagrammatic representation of the given problem:
Assume x and y packets be transported from factory A to the agencies P and Q respectively.
After that,
[60 − (x + y)] packets be transported to the agency R.
First constraint ⇢ x, y ≥ 0 and
Second constraint ⇢ 60 − (x + y) ≥ 0
(x + y) ≤ 60
The requirement at agency P is 40 packets.
Since, x packets are transported from factory A,
Therefore, the remaining (40 − x) packets are transported from factory B.
Similarly,
(40 − y) packets are transported by B to Q and 50− [60 − (x + y)]
i.e. (x + y − 10) packets will be transported from factory B to agency R respectively.
Number of packets cannot be negative.
Hence,
Third constraint ⇢ 40 – x ≥ 0
=> x ≤ 40
Fourth constraint ⇢ 40 – y ≥ 0
=> y ≤ 40
Fifth constraint ⇢ x + y – 10 ≥ 0
=> x + y ≥ 10
Thus,
Costs of transportation of each packet from factory A to agency P, Q, R are Rs 5, 4, 3.
Costs of transportation of each packet from factory B to agency P, Q, R are Rs 4, 2, 5.
Let total cost of transportation be Z.
Z = 5x + 4y + 3[60 – x + y] + 4(40 – x) + 2(40 – y) + 5(x + y – 10]
= 3x + 4y + 10
Hence,
The required mathematical formulation of linear programming is:
Minimize Z = 3x + 4y + 370
subject to constraints,
x + y ≤ 60
x ≤ 40
y ≤ 40
x + y ≥ 10
where, x, y ≥ 0
Let us convert inequations into equations as follows:
x + y = 60, x = 40, y = 40, x + y = 10, x = 0 and y = 0
Region represented by x + y ≤ 60:
The line x + y = 60 meets the coordinate axes at A_{1} (60, 0) and B_{1} (0, 60) respectively.
Region containing origin represents x + y ≤ 60 as (0, 0) satisfies x + y ≤ 60.
Region represented by x ≤ 40:
The line x = 40 is parallel to y-axis, meets x-axis at A_{2} (40, 0). Region containing origin represents x ≤ 40 as (0, 0) satisfies x ≤ 40.
Region represented by y ≤ 40:
The line y = 40 is parallel to x-axis, meets y-axis at B_{2} (0, 40).
Region containing origin represents y ≤ 40 as (0, 0) satisfies y ≤ 40.
Region represented by x + y ≥ 10:
The line x + y = 10 meets the coordinate axes at A_{2} (10, 0) and B_{3} (0, 10) respectively.
Region not containing origin represents x + y ≥ 10 as (0, 0) does not satisfy x + y ≥ 10.
Shaded region A_{3} A_{2} P Q B_{2} B_{3} represents feasible region.
Point P(40, 20) is obtained by solving x = 40 and x + y = 60
Point Q(20, 40) is obtained by solving y = 40 and x + y = 60
The value of Z = 3x + 4y + 370 at
A_{3}(10, 0) = 3(10) + 4(0) + 370 = 400
A_{2}(40, 0) = 3(40) + 4(0) + 370 = 490
P(40, 20) = 3(40) + 4(20) + 370 = 570
Q(20, 40) = 3(20) + 4(40) + 370 = 590
B_{2}(0, 40) = 3(0) + 4(40) + 370 = 530
B_{3}(0, 10) = 3(0) + 4(10) + 370 = 410
Hence, minimum value of Z = 400 at x = 10, y = 0
So,
From A -> P = 10 packets
From A -> Q = 0 packets
From A -> R = 50 packets
From B -> P = 30 packets
From B -> Q = 40 packets
From B -> R = 0 packets
Therefore, minimum cost = Rs 400