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

 How many packets from each factory be transported to each agency so that the cost of transportation is minimum? Also find the minimum cost?

Answer 1

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,

\[Z = 5x + 4y + 3\left[ 60 - \left( x + y \right) \right] + 4\left( 40 - x \right) + 2\left( 40 - y \right) + 5\left( x + y - 10 \right)\]
\[ = 3x + 4y + 10\]

Minimize Z =   \[5x + 4y + 3\left( 60 - x - y \right) + 4\left( 40 - x \right) + 2\left( 40 - y \right) + 5\left( x + y - 10 \right)\]

= \[3x + 4y + 370\]

subject to

\[x + y \leq 60\]

\[x \leq 40\]

\[y \leq 40\]

\[x + y \geq 10\]

\[x, y \geq 0\]

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 corner points are D₁(0, 40), H₁(20, 40), G₁(40, 20), C₁(40, 0), E₁(10, 0) and F₁(0, 10). 
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​