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,

$$\begin{gathered} x\ge 0,y\ge 0\mathrm{and}60-x-y\ge 0 \\ \Rightarrow x\ge 0,y\ge 0\mathrm{and}x+y\le 60 \\ 40-x\ge 0,40-y\ge 0\mathrm{and}x+y-10\ge 0 \\ \Rightarrow x\mathit{\le }40,y\le 40\mathrm{and}x+y\ge 10 \end{gathered}$$

Total transportation cost Z is given by,

$$\begin{gathered} \mathrm{Z}=5x+4y+3\left[60-\left(x+y\right)\right]+4\left(40-x\right)+2\left(40-y\right)+5\left(x+y-10\right) \\ =3x+4y+10 \end{gathered}$$

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

$$\begin{gathered} x+y\le 60 \\ x\le 40 \\ y\le 40 \\ x+y\ge 10 \\ x,y\ge 0 \end{gathered}$$

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​