# RD Sharma Solutions Class 12 Linear Programming Exercise 30.5

RD Sharma Solutions for Class 12 Maths Chapter 30 Linear Programming Exercise 30.5, is given here. Highly experienced subject experts at BYJUâ€™S having vast knowledge of concepts have developed the solutions, which match the understanding ability of the students. To achieve a good academic score in Mathematics, students are required to practice all the questions given in this exercise. Students can easily download the RD Sharma Solutions pdf consisting of this chapter solutions, which are available in the links provided below.

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EXERCISE 30.5

Q1.

Solution:

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

Then, (100 â€“ x â€“ y) will be supplied to shop F.

The requirement at shopÂ DÂ is 60 quintals since,Â xÂ quintals are transported from godownÂ A.

Therefore, the remaining (60 âˆ’Â x) quintals will be transported from godownÂ B.

Similarly, (50 âˆ’Â y) quintals and 40 âˆ’ (100 âˆ’Â xÂ âˆ’Â y) i.e. (xÂ +Â yÂ âˆ’ 60) quintals will be transported from godownÂ BÂ to shopÂ EÂ andÂ FÂ respectively.

Here is the diagrammatic representation of the given problem:

x â‰¥ 0 , y â‰¥ 0 and 100 – x – y â‰¥ 0Â

â‡’ x â‰¥ 0 , y â‰¥ 0 , and x + y â‰¤ 100

60 – x â‰¥ 0 , 50 – y â‰¥ 0 , and x + y – 60 â‰¥ 0

â‡’ x â‰¤ 60 , y â‰¤ 50 , and x + y â‰¥ 60Â

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

Hence, the required mathematical formulation of linear programming is:

Minimize Z = 2.5xÂ + 1.5yÂ + 410Â

subject to the constraints,

x + y â‰¤ 100

x â‰¤ 60

y â‰¤ 50

x + y â‰¥ 60

x, y â‰¥ 0

The feasible region obtained by the system of constraints is:

The corner points areÂ A(60, 0),Â B(60, 40),Â C(50, 50), andÂ D(10, 50).

The values of Z at these corner points are as follows.

 Corner point Â Z = 2.5xÂ + 1.5yÂ + 410 A (60, 0) 560 B (60, 40) 620 C (50, 50) 610 D (10, 50) 510 -> minimum

Hence, the minimum value of Z is 510 atÂ D (10, 50).

Thus, the amount of grain transported fromÂ AÂ toÂ D,Â E, andÂ FÂ is 10 quintals, 50 quintals, and 40 quintals respectively.

FromÂ BÂ toÂ D,Â E, andÂ FÂ is 50 quintals, 0 quintals, and 0 quintals respectively.

The minimum cost is Rs 510.

Q2.

Solution:

The diagrammatic representation of the given problem:

LetÂ xÂ andÂ yÂ packets be transported from factoryÂ AÂ to the agenciesÂ PÂ andÂ QÂ respectively.Â

Then, [60 âˆ’ (xÂ +Â y)] packets be transported to the agencyÂ R.

First constraint:

x, y â‰¥ 0 and

Second constraint:

60 âˆ’ (xÂ +Â y) â‰¥ 0

(xÂ +Â y) â‰¤ 60

The requirement at agencyÂ PÂ is 40 packets. Since,Â xÂ packets are transported from factoryÂ A,Â

Therefore, the remaining (40 âˆ’Â x) packets are transported from factoryÂ B.

Similarly, (40 âˆ’Â y) packets are transported by B to Q and 50âˆ’Â [60 âˆ’ (xÂ +Â y)] i.e. (xÂ +Â yÂ âˆ’Â 10) packets will be transported from factoryÂ BÂ to agencyÂ RÂ respectively.

Number of packets cannot be negative.

Therefore,

Third constraint:

40 – x â‰¥ 0

=> x â‰¤ 40

Fourth constraint:

40 – y â‰¥ 0

=> y â‰¤ 40

Fifth constraint:

x + y – 10 â‰¥ 0

=> x + y â‰¥ 10

So, 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 =Â 3x + 4y + 370

subject to constraints,

x + y â‰¤ 60

x â‰¤ 40

y â‰¤ 40

x + y â‰¥ 10

where, x, y â‰¥ 0

Let us convert inequations into equations as follows:

xÂ +Â yÂ = 60,Â xÂ = 40,Â yÂ = 40,Â xÂ +Â yÂ = 10,Â xÂ = 0 andÂ yÂ = 0

Region represented byÂ xÂ +Â yÂ â‰¤ 60:

The lineÂ xÂ +Â yÂ = 60 meets the coordinate axes atÂ A1(60, 0) andÂ B1(0, 60) respectively. Region containing origin represents xÂ +Â yÂ â‰¤ 60 as (0, 0) satisfies xÂ +Â yÂ â‰¤ 60.

Region represented byÂ x â‰¤ 40:

The lineÂ x = 40 is parallel to y-axis, meets x-axis atÂ A2(40, 0). Region containing origin represents x â‰¤ 40 as (0, 0) satisfies x â‰¤ 40.

Region represented byÂ y â‰¤ 40:

The lineÂ y = 40 is parallel to x-axis, meets y-axis atÂ B2(0, 40). Region containing origin represents y â‰¤ 40 as (0, 0) satisfies y â‰¤ 40.

Region represented byÂ x + y â‰¥ 10:

The lineÂ x + y = 10 meets the coordinate axes atÂ A2(10, 0) andÂ B3(0, 10) respectively. Region not containing origin represents x + y â‰¥ 10 as (0, 0) does not satisfy x + y â‰¥ 10.

Shaded region A3 A2 P Q B2 B3 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

A3(10, 0) = 3(10) + 4(0) + 370 = 400

A2(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

B2(0, 40) = 3(0) + 4(40) + 370 = 530

B3(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