The given information can be exhibited diagrammatically as follows:

Let the factory at *P* transports *x* units of commodity to depot at *A* and *y*units to depot at *B*. Since the factory at *P* has the capacity of 8 units of the commodity. Therefore, the left out (8 – *x* – *y*) units will be transported to depot at *C*.

Since the requirements are always non-negative quantities. Therefore,

*x* ≥ 0, *y* ≥ 0 and 8 – *x* – *y* ≥ 0 ⇒ *x* ≥ 0, *y* ≥ 0 and *x* + *y* ≤ 8

Since the weekly requirement of the depot at *A* is 5 units of the commodity and *x* units are transported from the factory at *P*. Therefore, the remaining (5 – *x*) units are to be transported from the factory at *Q*. Similarly, 5 – *y*units of the commodity will be transported from the factory at *Q* to the depot at *B*. But the factory at *Q* has the capacity of 6 units only, therefore the remaining 6 – (5 – *x* + 5 – *y*) = *x* + *y* – 4 units will be transported to the depot at *C*. As the requirements at the depots at *A*, *B* and *C* are always non-negative.

∴ 5 – *x* ≥ 0, 5 – *y* ≥ 0 and *x* + *y* – 4 ≥ 0

⇒ *x* ≤ 5, *y* ≤ 5 and *x* + *y* ≥ 4.

The transportation cost from the factory at *P* to the depots at *A*, *B* and *C*are respectively Rs 16*x*, 10*y* and 15(8 – *x* – *y*). Similarly, the transportation cost from the factory at *Q* to the depots at *A*, *B* and *C* are respectively Rs 10(5 – *x*), 12(5 – *y*) and 10(*x* + *y* – 4). Therefore, the total transportation cost *Z* is given by

*Z* = 16*x* + 10*y* + 15(8 – *x* – *y*) + 10(5 – *x*) + 12(5 – *y*) + 10(*x* + *y* – 4)

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

Find *x* and *y* which

Minimize *Z* = *x* – 7*y* + 190

Subject to

*x* + *y* ≤ 8

*x* + *y* ≥ 4

*x* ≤ 5

*y* ≤ 5

and, *x* ≥ 0, *y* ≥ 0