The given information can be exhibited diagrammatically as shown in fig.
Let the depot A transport x thousands bricks to builders P, y thousands to builder Q. Since the depot A has stock of 30,000 bricks. Therefore, the remaining bricks i.e. 30 – (x + y) thousands bricks will be transported to the builder R.
Since the number of bricks is always a non-negative real number.
Therefore,
x ≥ 0, y ≥ 0 and 30 – (x + y) ≥ 0 ⇒ x ≥ 0, y ≥ 0 and x + y ≤ 30.
Now, the requirement of the builder P is of 15000 bricks and x thousand bricks are transported from the depot A. Therefore, the remaining (15 – x) thousands bricks are to be transported from the depot at B. The requirement of the builder Q is of 20,000 bricks and y thousand bricks are transported from depot A. Therefore, the remaining (20 – y) thousand bricks are to be transported from depot B.
Now, depot B has 20 – (15 – x +20 – y) = x + y – 15 thousand bricks which are to be transported to the builder R.
Also, 15 – x ≥ 0, 20 – y ≥ 0 and x + y – 15 ≥ 0
⇒ x ≤ 15, y ≤ 20 and x + y ≥ 15
The transportation cost from the depot A to the builders P, Q and R are respectively Rs 40x, 20y and 30(30 – x – y). Similarly, the transportation cost from the depot B to the builders P, Q and R are respectively Rs 20 (15 – x), 60(20 – y) and 40(x + y – 15) respectively. Therefore, the total transportation cost Z is given by
Z = 40x + 20y + 30(30 – x – y) + 20(15 – x) + 60(20 – y) + 40(x + y – 15)
⇒ Z = 30x – 30y + 1800
Hence, the above LLP can be stated mathematically as follows:
Find x and y in thousands which
Minimize Z = 30x – 30y + 1800
Subject to
x + y ≤ 30
x ≤ 15
y ≤ 20
x + y ≥ 15
and, x ≥ 0, y ≥ 0
