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 40*x*, 20*y* 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* = 40*x* + 20*y* + 30(30 – *x* – *y*) + 20(15 – *x*) + 60(20 – *y*) + 40(*x* + *y* – 15)

⇒ *Z* = 30*x* – 30*y* + 1800

Hence, the above LLP can be stated mathematically as follows:

Find *x* and *y* in thousands which

Minimize *Z* = 30*x* – 30*y* + 1800

Subject to

*x* + *y* ≤ 30

*x* ≤ 15

*y* ≤ 20

*x* + *y* ≥ 15

and, *x* ≥ 0, *y* ≥ 0