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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 30 |

Exercise | 30.5 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 30.5 Solutions Chapter 30 Linear Programming**

# Class 12 RD Sharma Solutions – Chapter 30 Linear Programming – Exercise 30.5

**Question 1. Two godowns A and B have gain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:**

Transportation cost per quintal (in Rs) | ||
---|---|---|

From/ To | A | B |

D | 6 | 4 |

E | 3 | 2 |

F | 2.50 | 3 |

**How should the supplies be transported in order that the transportation cost is minimum? what is the minimum cost?**

**Solution:**

Let us assume godown A supply x and y quintals of grain to the shops D and E.

Afterwards, (100 – x – y) will be supplied to shop F.

The requirement at shop D is 60 quintals

Since, x quintals are transported from godown A.

Therefore, the remaining (60 − x) quintals will be transported from godown B.

Similarly,

(50 − y) quintals and 40 − (100 − x − y) i.e. (x + y − 60) quintals will be transported from godown B to shop E and F respectively.

The diagrammatic representation of the given problem:

x ≥ 0 , y ≥ 0 and 100 – x – y ≥ 0

⇒ x ≥ 0 , y ≥ 0 , and x + y ≤ 100

60 – x ≥ 0 , 50 – y ≥ 0 , and x + y – 60 ≥ 0

⇒ x ≤ 60 , y ≤ 50 , and x + y ≥ 60

Total transportation cost Z is given by,

Z = 6x + 3y + 2.5(100 – x – y) + 4(60 – x) + 2(50 – y) + 3( x + y – 60)

= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x + 100 – 2y + 3x + 3y – 180

= 2.5x + 1.5y + 410

Thus, the required mathematical formulation of linear programming is:

Minimize Z = 2.5x + 1.5y + 410

subject to the constraints,

x + y ≤ 100

x ≤ 60

y ≤ 50

x + y ≥ 60

x, y ≥ 0

The feasible region obtained by the system of constraints is:

The corner points are A(60, 0), B(60, 40), C(50, 50), and D(10, 50).

The values of Z at these corner points are given below.

Corner point | Z = 2.5x + 1.5y + 410 |

A (60, 0) | 560 |

B (60, 40) | 620 |

C (50, 50) | 610 |

D (10, 50) | 510 -> minimum |

Therefore,

The minimum value of Z is 510 at D (10, 50).

Hence, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively.

From B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.

The minimum cost is Rs 510.

**Question 2. A medical company has factories at two places, A and B. From these places, supply is made to each of its three agencies situated at P, Q and R. The monthly requirements of the agencies are respectively 40, 40 and 50 packets of the medicines, while the production capacity of the factories, A and B are 60 and 70 packets respectively. The transportation cost per packet from the factories to the agencies are given below:**

Transportation cost per packet (in Rs) | ||

To\From | A | B |

P | 5 | 4 |

Q | 4 | 2 |

R | 3 | 5 |

**How many packets from each factory be transported to each agency so that the cost of transportation is minimum? Also**,** find the minimum cost?**

**Solution:**

The diagrammatic representation of the given problem:

Assume x and y packets be transported from factory A to the agencies P and Q respectively.

After that,

[60 − (x + y)] packets be transported to the agency R.

First constraint ⇢ x, y ≥ 0 and

Second constraint ⇢ 60 − (x + y) ≥ 0

(x + y) ≤ 60

The requirement at agency P is 40 packets.

Since, x packets are transported from factory A,

Therefore, the remaining (40 − x) packets are transported from factory B.

Similarly,

(40 − y) packets are transported by B to Q and 50− [60 − (x + y)]

i.e. (x + y − 10) packets will be transported from factory B to agency R respectively.

Number of packets cannot be negative.

Hence,

Third constraint ⇢ 40 – x ≥ 0

=> x ≤ 40

Fourth constraint ⇢ 40 – y ≥ 0

=> y ≤ 40

Fifth constraint ⇢ x + y – 10 ≥ 0

=> x + y ≥ 10

Thus,

Costs of transportation of each packet from factory A to agency P, Q, R are Rs 5, 4, 3.

Costs of transportation of each packet from factory B to agency P, Q, R are Rs 4, 2, 5.

Let total cost of transportation be Z.

Z = 5x + 4y + 3[60 – x + y] + 4(40 – x) + 2(40 – y) + 5(x + y – 10]

= 3x + 4y + 10

Hence,

The required mathematical formulation of linear programming is:

Minimize Z = 3x + 4y + 370

subject to constraints,

x + y ≤ 60

x ≤ 40

y ≤ 40

x + y ≥ 10

where, x, y ≥ 0

Let us convert inequations into equations as follows:

x + y = 60, x = 40, y = 40, x + y = 10, x = 0 and y = 0

Region represented by x + y ≤ 60:

The line x + y = 60 meets the coordinate axes at A_{1} (60, 0) and B_{1} (0, 60) respectively.

Region containing origin represents x + y ≤ 60 as (0, 0) satisfies x + y ≤ 60.

Region represented by x ≤ 40:

The line x = 40 is parallel to y-axis, meets x-axis at A_{2} (40, 0). Region containing origin represents x ≤ 40 as (0, 0) satisfies x ≤ 40.

Region represented by y ≤ 40:

The line y = 40 is parallel to x-axis, meets y-axis at B_{2} (0, 40).

Region containing origin represents y ≤ 40 as (0, 0) satisfies y ≤ 40.

Region represented by x + y ≥ 10:

The line x + y = 10 meets the coordinate axes at A_{2} (10, 0) and B_{3} (0, 10) respectively.

Region not containing origin represents x + y ≥ 10 as (0, 0) does not satisfy x + y ≥ 10.

Shaded region A_{3} A_{2} P Q B_{2} B_{3} represents feasible region.

Point P(40, 20) is obtained by solving x = 40 and x + y = 60

Point Q(20, 40) is obtained by solving y = 40 and x + y = 60

The value of Z = 3x + 4y + 370 at

A_{3}(10, 0) = 3(10) + 4(0) + 370 = 400

A_{2}(40, 0) = 3(40) + 4(0) + 370 = 490

P(40, 20) = 3(40) + 4(20) + 370 = 570

Q(20, 40) = 3(20) + 4(40) + 370 = 590

B_{2}(0, 40) = 3(0) + 4(40) + 370 = 530

B_{3}(0, 10) = 3(0) + 4(10) + 370 = 410

Hence, minimum value of Z = 400 at x = 10, y = 0

So,

From A -> P = 10 packets

From A -> Q = 0 packets

From A -> R = 50 packets

From B -> P = 30 packets

From B -> Q = 40 packets

From B -> R = 0 packets

Therefore, minimum cost = Rs 400

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