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Question
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.) | ||
| From-> | A | B |
| To | ||
| P | 5 | 4 |
| Q | 4 | 2 |
| R | 3 | 5 |
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Solution
Let x and y packets be transported from factory A to the agencies P and Q respectively. Then, [60 − (x + y)] packets be transported to the agency R.
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.Therefore,
\[x \geq 0, y \geq 0 \text{ and } 60 - x - y \geq 0\]
\[ \Rightarrow x \geq 0, y \geq 0 \text{ and } x + y \leq 60\]
\[40 - x \geq 0, 40 - y \geq 0 \text{ and } x + y - 10 \geq 0\]
\[ \Rightarrow x \leq 40, y \leq 40 \text{ and } x + y \geq 10\]
Total transportation cost Z is given by,
\[Z = 5x + 4y + 3\left[ 60 - \left( x + y \right) \right] + 4\left( 40 - x \right) + 2\left( 40 - y \right) + 5\left( x + y - 10 \right)\]
\[ = 3x + 4y + 10\]
\[x + y \leq 60\]
\[x \leq 40\]
\[y \leq 40\]
\[x + y \geq 10\]
\[x, y \geq 0\]
First we will 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 A1(60, 0) and B1(0, 60) respectively. By joining these points we obtain the line x + y = 60. Clearly (0,0) satisfies the x + y = 60. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 60.
Region represented by x ≤ 40:
x = 40 is the line that passes C1(40, 0) and is parallel to the Y axis.The region to the left of the line x = 40 will satisfy the inequation x ≤ 40.
Region represented by y ≤ 40:
y = 40 is the line that passes D1(0, 40) and is parallel to the X axis . The region below the line y = 40 will satisfy the inequation y ≤ 40.
Region represented by x + y ≥ 10:
The line x + y = 10 meets the coordinate axes at E1(10, 0) and \[F_1 \left( 0, 10 \right)\] respectively. By joining these points we obtain the line x + y = 10. Clearly (0,0) does not satisfies the inequation x + y ≥ 10. So,the region which does not contain the origin represents the solution set of the inequation x + y ≥ 10.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 60, x ≤ 40, y ≤ 40, x + y ≥ 10, x ≥ 0 and y ≥ 0 are as follows.
The values of Z at these corner points are as follows
| Corner point | Z= 3x + 4y + 370 |
| D1 | 530 |
| H1 | 590 |
| G1 | 570 |
| C1 | 490 |
| E1 | 400 |
| F1 | 410 |
The minimum value of Z is 400 which is at E1(10, 0).
Thus, the minimum cost is Rs 400.
Hence,
From A: 10 packets, 0 packets and 50 packets to P, Q and R respectively
From B: 30 packets, 40 packets and 0 packets to P, Q and R respectively
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