Maharashtra Board Class 12 Maths Part 2 Chapter 6 Linear Programming 6.1 Solutions

Question 1. A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to a machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B and the number of man-hours available for the firm is as follows.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका दी गई है जो गैजेट्स A और B के लिए फाउंड्री और मशीन शॉप में आवश्यक मैन-आवर्स दर्शाती है। गैजेट A के लिए फाउंड्री में 10 घंटे और मशीन शॉप में 5 घंटे लगते हैं, जबकि गैजेट B के लिए फाउंड्री में 6 घंटे और मशीन शॉप में 4 घंटे लगते हैं। फाउंड्री के लिए कुल 60 घंटे और मशीन शॉप के लिए 35 घंटे उपलब्ध हैं।
Profit on the sale of A is Rs. 30 and B Rs. 20 Per unit Formulate the LPP to have maximum profit.
Answer: Solution:
Let the manufacturing firm produce x gadgets of type A and y gadgets of type B.
On selling x gadgets of type A the firm gets Rs. 30 and that on type B is Rs. 20.
. Total profit is z = Rs. 30x + 20y.
Since x and y are the numbers of gadgets, x ≥ 0, y ≥ 0
From the given table, the availability of man-hours of labour required in each shop and for the firm is given as 60 and 35.
. The inequation are 10x + 6y ≤ 60 and 5x + 4y ≤ 35.
Hence the given LPP can be formulated as Maximize z = 30x + 20y
Subject to 10x + 6y ≤ 60, 5x + 4y ≤ 35, x ≥ 0, y ≥ 0.
In simple words: This problem involves setting up a Linear Programming Problem to maximize profit from producing two types of gadgets, A and B, considering man-hour constraints in two departments (foundry and machine shop) and individual profit margins for each gadget.

🎯 Exam Tip: Clearly define decision variables (x and y) and accurately translate profit information into the objective function. Constraints derived from resource availability (man-hours) must be formulated correctly with appropriate inequality signs.

Question 2. In a cattle breeding farm, it is prescribed that the food ratio for one animal must contain 14, 22, and 1 unit of nutrients A, B, and C respectively. Two different kinds of fodder are available. Each unit weight of these two contains the following:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका पोषक तत्वों A, B, C की मात्रा को दर्शाती है जो विभिन्न प्रकार के चारे में मौजूद हैं। चारा 1 में 2 इकाई पोषक तत्व A, 2 इकाई पोषक तत्व B और 1 इकाई पोषक तत्व C है। चारा 2 में 1 इकाई पोषक तत्व A, 3 इकाई पोषक तत्व B और 1 इकाई पोषक तत्व C है।
The cost of fodder 1 is Rs. 3 per unit and that of fodder 2 is Rs. 2 per unit. Formulate the LPP to minimize the cost.
Answer: Solution:
Let x unit of fodder 1 and y unit of fodder 2 be included in the ration of an animal
The cost of 1 unit of fodder 1 is Rs. 3 and the cost of 1 unit of fodder 2 is Rs. 2.
. The total cost is 3x + 2y.
The minimum requirement of the nutrients A, B, and C is given as 14 units, 22 units, and 1 unit.
. From the given table, the daily food ration will include (2x + 2y) unit of Nutrient A, (2x + 3y) unit of Nutrient B, and (x + y) of Nutrient C.
The total cost is 2 = 3x + 2y
Hence the given LPP can be formulated as Minimize z = 3x + 2y
subject to 2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1, x ≥ 0, y ≥ 0.
In simple words: This problem involves formulating an LPP to minimize the cost of an animal's food ration while ensuring minimum required nutrient levels (A, B, and C) are met from two available types of fodder.

🎯 Exam Tip: Pay close attention to minimum requirements (≥) versus maximum availability (≤) for constraints. Ensure the objective function correctly reflects the cost to be minimized.

Question 3. A Company manufactures two types of chemicals A and B. Each chemical requires two types of raw materials P and Q. The table below shows a number of units of P and Q required to manufacture one unit of A and one unit of B.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका रसायनों A और B के उत्पादन के लिए आवश्यक कच्चे माल P और Q की इकाइयों को दर्शाती है। रसायन A को बनाने के लिए 3 इकाई P और 2 इकाई Q की आवश्यकता होती है, जबकि रसायन B को बनाने के लिए 2 इकाई P और 5 इकाई Q की आवश्यकता होती है। कच्चे माल P की 120 इकाई और Q की 160 इकाई उपलब्ध हैं।
The company gets profits of Rs. 350/- and Rs. 400/- by selling one unit of A and one unit of B respectively. Formulate the problem as LPP to maximize the profit.
Answer: Solution:
. Let the company manufactures x unit of chemical A and y unit of chemical B.
The availability of the raw materials for the production of chemicals A and B are given as 120 and 160 units.
The company gets Rs. 350 as profit on selling one unit of chemical A and Rs. 400 as profit on selling one unit of chemical B.
. Total profit is Rs. (350x + 400y).
The inequation can be written as.
3x + 2y ≤ 120
2x + 5y ≤ 160
and x & y cannot be negative
Hence the LPP can be formulated as follows,
Maximize z = 350x + 400y
Subject to 3x + 2y ≤ 120, 2x + 5y ≤ 160, x ≥ 0, y ≥ 0.
In simple words: This problem requires setting up an LPP to maximize profit by determining the optimal production quantities of two chemicals (A and B), given constraints on the availability of two raw materials (P and Q) and per-unit profit for each chemical.

🎯 Exam Tip: When formulating profit maximization problems, ensure the objective function correctly sums up the profits from each product and constraints accurately reflect raw material limitations (usually ≤).

Question 4. A printing company prints two types of magazines A and B. The company earns Rs. 10 and Rs. 15 on magazines A and B per copy. These are processed on three machines I, II, III. Magazine A requires 2 hours on the machine I, 5 hours on machine II, and 2 hours on machine III. Magazine B requires 3 hours on machine 1, 2 hours on machine II and 6 hours on machine III. Machines I, II, III are available for 36, 50, 60 hours per week respective. Formulate the linear programming problem to maximize the profit.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका मशीनों I, II, और III पर पत्रिका A और B के लिए आवश्यक समय को दर्शाती है। पत्रिका A को मशीन I पर 2 घंटे, मशीन II पर 5 घंटे और मशीन III पर 2 घंटे लगते हैं, जबकि पत्रिका B को मशीन I पर 3 घंटे, मशीन II पर 2 घंटे और मशीन III पर 6 घंटे लगते हैं। मशीनों I, II, III के लिए क्रमशः 36, 50, और 60 घंटे प्रति सप्ताह उपलब्ध हैं।
Answer: Solution:
Let the company print x magazine of type A and y magazines of type B.
Then the total earnings of the company are Rs. 10x + 15y.
The given problem can be tabulated as follows.
From the table, the total time required for Machine I is (2x + 3y) hours, for machine II is (5x + 2y) hours, and for machine III is (2x + 6y) hours.
The machine I, II, and III are available for 36, 50, and 60 hours per work.
. The constraints are 2x + 3y ≤ 36, 5x + 2y ≤ 50 and 2x + 6y ≤ 60.
Since x and y cannot be negative, we have x ≥ 0, y ≥ 0.
Hence the given LPP can be formulated as
Maximize z = 10x + 15y
Subject to 2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.
In simple words: This problem involves creating an LPP to maximize a printing company's profit by deciding the number of two types of magazines to print, considering the time constraints on three different machines and the profit earned per magazine.

🎯 Exam Tip: Pay attention to converting per-unit machine time requirements and total machine availability into correct inequality constraints. The objective function should directly reflect the total profit from both magazine types.

Question 5. Manufacture produces bulbs and tubes. Each of these must be processed through two machines M₁ and M2. A package of bulbs requires 1 hour of work on machine M₁ and 3 hours of work on M2. A package of tubes requires 2 hours on machine M₁ and 4 hours on machine M2. He earns a profit of Rs. 13.5 per package of bulbs and Rs. 55 per package of tubes. Formulate the LPP to maximize the profit. He operates M₁ for at most 10 hours and M2 for at most 12 hours a day.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका मशीनों M₁ और M₂ पर बल्ब और ट्यूब के पैकेजों के लिए आवश्यक समय को दर्शाती है। बल्ब के एक पैकेज को M₁ पर 1 घंटा और M₂ पर 3 घंटे लगते हैं, जबकि ट्यूब के एक पैकेज को M₁ पर 2 घंटे और M₂ पर 4 घंटे लगते हैं। मशीन M₁ के लिए 10 घंटे और मशीन M₂ के लिए 12 घंटे उपलब्ध हैं।
Answer: Solution:
Let the manufacturer produce x packages of bulbs and y packages of tubes.
He earns a profit of Rs. 13.5 per packages of bulbs and Rs. 55 per package of tubes.
. His total profit = Rs. (13.5x + 55y).
The given problem can be tabulated as follows.
From the above table, the total time required for M₁ is (x + 2y), and that of M2 is (3x + 4y).
M₁ and M2 are available for at most 10 hrs per day and 12 hours per day.
. The constraint for the objective function is x + 2y ≤ 10, 3x + 4y ≤ 12
Hence the give LPP can be formulated as
Maximize z = 13.5x + 55y
Subject to x + 2y ≤ 10, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0.
In simple words: This problem focuses on creating an LPP to maximize a manufacturer's profit by deciding the number of bulb and tube packages to produce, given processing time limits on two machines and the profit for each package type.

🎯 Exam Tip: Differentiate between "at most" (≤) and "at least" (≥) when forming constraints based on machine availability. Ensure fractional profit values are handled correctly in the objective function.

Question 6. A Company manufactures two types of fertilizers F₁ and F2. Each type of fertilizer requires two raw materials A and B. The number of units of A and B required to manufacture one unit of fertilizer F₁ and F2 and availability of the raw materials A and B per day are given in the table below
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका उर्वरक F₁ और F₂ के उत्पादन के लिए आवश्यक कच्चे माल A और B की इकाइयों को दर्शाती है। उर्वरक F₁ को बनाने के लिए 2 इकाई A और 1 इकाई B की आवश्यकता होती है, जबकि उर्वरक F₂ को बनाने के लिए 3 इकाई A और 4 इकाई B की आवश्यकता होती है। कच्चे माल A की 40 इकाई और B की 70 इकाई प्रतिदिन उपलब्ध हैं।
By selling one unit of F₁ and one unit of F2, the company gets a profit of Rs. 500 and Rs. 750 respectively. Formulate the problem as LPP to maximize the profit.
Answer: Solution:
Let the company manufacture x units of Fertilizers F₁ and y units of fertilizer F2.
The company gets a profit of Rs. 500 and Rs. 750 by selling a unit of F₁ and F2.
. Total profit = Rs. (500x + 750y)
The availability of raw materials A and B per day is given as 40 and 70.
. From the given table the constraints can be written as 2x + 3y ≤ 40 and x + 4y ≤ 70.
Since x & y cannot be negative, x ≥ 0, y ≥ 0
Hence the given LPP can be formulated as
Maximize z = 500x + 750y
Subject to 2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0.
In simple words: This problem involves formulating an LPP to maximize profit by deciding the production quantities of two types of fertilizers (F₁ and F₂), given the availability of two raw materials (A and B) and the profit from selling each fertilizer type.

🎯 Exam Tip: Accurately interpret the raw material requirements for each product from the table to form the constraint inequalities. Ensure the objective function reflects the total profit contribution from both fertilizers.

Question 7. A doctor has prescribed two different kinds of feeds A and B to form a weekly diet for a sick person. The minimum requirement of fats, carbohydrates, and proteins are 18, 28,14 units respectively. One unit of food A has 4 units of fat, 14 units of carbohydrates, and 8 units of protein. One unit of food B has 6 units of fat, 12 units of carbohydrates and 8 units of protein. The price of food A is Rs. 4.5 per unit and that of food B is Rs. 3.5 per unit. Form the LPP so that the sick person's diet meets the requirements at minimum cost.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका पोषक तत्वों वसा, कार्बोहाइड्रेट और प्रोटीन की मात्रा को दर्शाती है जो भोजन A और B में मौजूद हैं, साथ ही न्यूनतम आवश्यकताएं भी। भोजन A में 4 इकाई वसा, 14 इकाई कार्बोहाइड्रेट और 8 इकाई प्रोटीन है, जबकि भोजन B में 6 इकाई वसा, 12 इकाई कार्बोहाइड्रेट और 8 इकाई प्रोटीन है। न्यूनतम आवश्यकताएं वसा के लिए 18, कार्बोहाइड्रेट के लिए 28 और प्रोटीन के लिए 14 इकाइयां हैं।
Answer: Solution:
Let x unit of food A and y unit of food B be consumed by a sick person.
The cost of food A in 4.5 per unit and food B is 3.5 per unit.
. Total cost = Rs. (4.5x + 3.5y)
The given conditions can be tabulated as follows.
. The given LPP can be formulated as
Minimise z = 4.5x + 3.5y
Subject to 4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14, x ≥ 0, y ≥ 0.
In simple words: This problem involves setting up an LPP to minimize the cost of a sick person's diet by combining two types of food, A and B, to meet minimum daily requirements for fats, carbohydrates, and proteins.

🎯 Exam Tip: When dealing with dietary problems, ensure that all nutrient requirements are formulated as "at least" (≥) constraints. The objective function should accurately represent the total cost to be minimized.

Question 8. If John drives a car at a speed of 60 kms-hour he has to spend Rs. 5 per km on petrol. If he drives at a faster speed of 90 km- hour, the cost of petrol increases to Rs. 8 per km. He has Rs. 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as LPP.
Answer: Solution:
Let John drive x km at a speed of 60 km/hr and y km at a speed of 90 km/hr.
. Time required to drive a distance of x km is \(x/60\) hours and the time require to drive at a distance of y km is \(y/90\) hours.
. Total time required \( (x/60 + y/90) \) hours.
Since he wishes to drive maximum distance within an hour,
\(x/60 + y/90 \le 1\)
He has to spend Rs. 5 per km at a speed of 60 km/hr and Rs. 8 per km at a speed of 90 km/hr.
He has Rs. 600 on petrol to spend, 5x + 8y ≤ 600
The total distance he wishes to travel is (x + y) hours.
. The given LPP can be formulated as
Maximize z = x + y
Subject to
\(x/60 + y/90 \le 1\)
\(5x + 8y \le 600\)
\(x \ge 0\)
\(y \ge 0\)
In simple words: This problem asks us to create an LPP to maximize the total distance John can travel by driving at two different speeds, while staying within budget for petrol and a total travel time of one hour.

🎯 Exam Tip: Accurately calculate time and cost for each driving speed. Constraints for both time and budget must be carefully formulated using appropriate inequality signs.

Question 9. The company makes concrete bricks made up of cement and sand. The weight of a concrete brick has to be at least 5 kg. Cement costs Rs. 20 per kg. and sand costs Rs. 6 per kg. Strength considerations dictate that a concrete brick should contain a minimum of 4 kg of cement and not more than 2 kg of sand. Formulate the LPP for the cost to be minimum.
Answer: Solution:
Let the concrete brick contain x kg of cement and y kg of sand.
The cost of cement is Rs. 20 per kg and sand is Rs. 6 per kg.
. The total cost = Rs. (20x + 6y)
Since the weight of the concrete brick has to be at least 5 kg, therefore, x + y ≥ 5
Also, the concrete brick should contain a minimum of 4 kg of cement, i.e. x ≥ 4,
and not more than 2 kg of sand, i.e, y ≤ 2.
. The LPP can be formulated as
Minimize z = 20x + 6y
Subject to x + y ≥ 5, x ≥ 4, y ≤ 2, x ≥ 0, y ≥ 0.
In simple words: This problem requires formulating an LPP to minimize the cost of making a concrete brick by determining the optimal amounts of cement and sand, while adhering to total weight, minimum cement, and maximum sand requirements.

🎯 Exam Tip: Carefully distinguish between "at least" (≥) and "not more than" (≤) when setting up the constraints for minimum cement and maximum sand. The objective function should correctly represent the total cost.