# CBSE Class 12-science Answered

**There is a factory located at each of the two places P and Q. From these location, a certain commodity is delivered to each of these depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below : To From Cost (in Rs) A B C P 16 10 15 Q 10 12 10 How many units should be transported from each factory to each depot in order that the transportation cost in minimum. Formulate the above LPP mathematically and the solve it.**

Let the factory at P transports x units of commodity to depot at A and y units to depot at B. Then, the mathematical model to the LPP is as follows :

Now, the weekly requirement of the depot at A is 5 units of the commodity. Since x units are transported from the factory at P, the remaining (5 – x) units need to be transported from the factory at Q. Obviously, 5 – x ≥ 0, i.e. x ≤ 5.

Similarly,

(5 – y) and 6 – (5 – x + 5 – y) = x + y – 4 units are to be transported from

the factory at Q to the depots at B and C respectively.

Thus, 5 – y ≥ 0 , x + y – 4 ≥0

i.e. y ≤ 5 , x + y ≥ 4

Total transportation cost Z is given by :

Z = 16x + 10y +1 0(5-x) + 12(5-y) + 10(x+y-4) + 15(8-x-y)

= x - 7y + 190

Minimize Z = x - 7y + 190

Subject to x + y < 8

x + y < 4

x < 5

y < 5

and x > 0, y > 0

To solve this LPP graphically, we first convert the in equations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in Fig

The coordinates of the corner points of the feasible region A_{2} A_{3} PQ B_{3} B_{2} are A_{2} (4, 0), A_{3} (5, 0), P(5, 3), Q (3, 5), B_{3} (0, 5) and B_{2} (0, 4). These points have been obtained by solving the corresponding intersecting line simultaneously.

The values of the objective function at these points are given in the following table :

Point (x, y) Value of the objective function Z= x - 7y + 190 |

A A P (5, 3) Z = 5 - 7 x 3 + 190 = 174 Q (3, 5) Z = 3 - 7 x 5 + 190 = 158 B B |

Clearly, Z is minimum at x = 0, y = 5. The minimum value of Z is 155.

Thus, the optimal transportation strategy will be to deliver 0, 5 and 3 units from the factory at P and 5, 0 and 1 unit from the factory at Q to the depots at A, B and C respectively. The minimum transportation cost in this case in Rs 155.