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  • Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 49

Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 49

Answers (1)

Answer: The minimum transportation cost is Rs.1550

Hint:

Use properties of LPP

Given:

From  To

Cost (in Rs.)

                                    

            A                         B                                C

P

           160                      100                             150

Q

           100                      120                             100

Solution:

Here, demand of the commodity (5 + 5 +4 = 14 units) is equal the supply of the commodity (8 + 6 = 14 units). So, no commodity could be left at the two factories.

Let x units and y units of the commodity be transported from the factory P to the depots A and B respectively.

Then (8-x-y) units of the commodity will be transported from the factory P to the depot C.

Now, the weekly requirement of depot A is 5 units of the commodity. Now, x units of the commodity are transported from factory P so the remaining (5-x) units of the commodity are transported from the factor Q to the depot A.

The weekly requirement of depot B is 5 units of the commodity. Now, y units of the commodity are transported from factory P. So the remaining (5-y) units of the commodity are transported from the factory Q to the depot B.

Similarly, 6-(5-x)-(5-y)=(x+y-4)   units of the commodity will be transported from the factory Q to the depot C.

Since the number of the units of commodity transported are from the factories to the depots are non-negative, therefore,

\begin{aligned} &x \geq 0, y \geq 0,8-x-y \geq 0,5-x \geq 0,5-y \geq 0, x+y-4 \geq 0 \\ & \end{aligned}

x \geq 0, y \geq 0, x+y \leq 8, x \leq 5, y \leq 5, x+y \geq 4

Total transportation cost =

\! \! \! \! \! \! \! \! \! 160 x+100 y+150(8-x-y)+100(5-x)+120(5-y)+100(x+y-4)=100 x-70 y+1900     

Thus, the given linear programming problem is

Minimize Z = 10x – 70y + 1900

Subject to constraints:

                                      \begin{aligned} &x+y \leq 8 \\ &x \leq 5 \\ &y \leq 5 \\ &x+y \geq 4 \\ &x \geq 0, y \geq 0 \end{aligned}

The feasible region determined by the given constraints can be diagrammatically represented as,

                              

The coordinates of the corner points of the feasible region are A(4,0),B(5,0),C(5,3), D(3,5), E(0,5) and F(0,4).

The value of the objective function at these points is given in the follow table.

Corner Points

Z=10x-70y+1900

(4,0)

10 \times 4-70 \times 0+1900=1940

(5,0)

10 \times 5-70 \times 0+1900=1950

(5,3)

10 \times 5-70 \times 3+1900=1740

(3,5)

10 \times 3-70 \times 5+1900=1580

(0,5)

10 \times 0-70 \times 5+1900=1550  (minimum)

(0,4)

10 \times 0-70 \times 4+1900=1620

 

The minimum value of Z is 1550 at x = 0, y = 5

Hence, for minimum transportation cost factory P should supply 0, 5, 3 units of commodity to depots A,B,C respectively and factory Q should supply 5,0,1 units of commodity to depots A,B and C respectively.

The minimum transportation cost is Rs.1550.

 

Posted by

infoexpert27

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