{"id":32718,"date":"2021-08-31T17:13:02","date_gmt":"2021-08-31T11:43:02","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=32718"},"modified":"2021-08-31T17:13:02","modified_gmt":"2021-08-31T11:43:02","slug":"ncert-solutions-for-class-12-maths-chapter-12-ex-12-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-12-ex-12-2\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 12 Linear Programming Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nReshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60\/kg and Food Q costs Rs 80\/kg. Food P contains 3 units\/ kg of Vitamin A and 5 units\/ kg of Vitamin B while food Q contains 4 units\/kg of Vitamin A and 2 units\/kg of vitamin B. Determine the minimum cost of the mixture.
\nSolution:
\nLet the mixture contain x units of food P and y units of food Q. We have the data as
\n\"NCERT
\nVitamin A constaint: 3x + Ay \u2265 8
\nVitamin B constraint: 5x + 2y \u2265 11
\nCost function : Z = 60x + 80y
\nThe L.P.P is Minimise Z = 60x + 80y
\nsubject to the constraints 3x + 4y \u22658, 5x + 2y \u2265 11, x, y \u2265 0.
\n\"NCERT
\nThe feasible region is shaded in the figure and is unbounded.
\nWe use comer point method to find the mini-mum of Z
\nFrom the table, the minimum value of Z is 160. Since the feasible region is unbounded j 160 may or may not be the minimum value i of Z. Consider the graph of the inequality 60x + 80y \u2264 160
\ni.e., 3x + 4y \u2264 8
\nThis half plane has no point in common with the feasible region. Then the minimum value of Z is 160 at the points on the line segment joining A\\(\\frac { 8 }{ 3 }\\), 0) and B(2, \\(\\frac { 1 }{ 2 }\\))<\/p>\n

Question 2.
\nOne kind of cake requires 200 g of flour and 25g of fat, and another kind of cake requires 100 g of flour and 50 g of fat Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients, used in making the cakes.
\nSolution:
\nLet x be the number of cakes of type I and y be the number of cakes of type II.
\nWe make use of the following table to write the L.P.P.
\nWe have the data as
\n\"NCERT
\nFlour constraint: 200x + 100y \u2264 5000
\ni.e., 2x + y \u2264 50
\nFat constraint: 25x + 50y \u2264 1000
\ni.e., x + 2y \u2264 40
\nTotal number of cakes : Z = x + y
\nThe L.P.P is Maximise Z = x + y
\nsubject to the constraints
\n2x + y \u2264 50, x + 2y \u2264 40, x, y >0.
\n\"NCERT
\n\u2234 Maximum value of Z is 30 at B(20, 10).
\nHence maximum number of cakes is 30, of which 20 are of type 1 and 10 are of type II.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nA factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman\u2019s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman\u2019s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman\u2019s time.
\ni. What number of rackets and bats must be made if the factory is to work at full capacity?
\nii. If the profit on a racket and on a bat is \u20b9 20 and \u20b9 10 respectively, find the maximum profit of the factory when it works at full capacity.
\nSolution:
\ni. Let x be the number of tennis rackets and that y be the cricket bats.
\n\"NCERT
\nMachine constraint: 1.5x + 3y \u2264 42
\ni.e., x + 2y \u2264 28
\nCraftman\u2019s constraint: 3x + y \u2264 24
\nProfit function : Z = 20x + 10y
\nThe L.P.P is Maximise Z = 20x + 10y subject to the constraints
\nx + 2y \u2264 28, 3x + y \u2264 24, x, y \u2265 0
\n\"NCERT
\n\u2234 Maximum value of Z is 200 at B(4, 12).
\nHence 4 tennis rackets and 12 cricket bats can be made if the factory is to work at full capacity.<\/p>\n

ii. The maximum profit is \u20b9 200\/-<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nA manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of \u20b9 17.50 per package on nuts and \u20b9 7.00 per package on bolts. How many package of each should be produced each day so as to maximise his profit, if he operates his machine for at the most 12 hours a day?
\nSolution:
\nLet x and y be the number of packages of nuts and bolts.
\n\u2234 We have the data as
\n\"NCERT
\nMachine A constraint: x + 3y \u2264 12
\nMachine B constraint: 3x + y \u2264 12
\nProfit function : Z = 17.5x + 7y
\nThe L.P.P is maximise Z = 17.5x + 7y subject to the constraints 1x + 3y \u2264 12, 3x+ 1y \u2264 12, x, y > 0
\n\"NCERT
\nMaximum value of Z is 73.50 at B(3, 3)
\nHence the manufacturer must produce 3 packages of nuts and 3 packages of bolts to maximise his profit. The maximum profit is \u20b9 73.50.<\/p>\n

Question 5.
\nA factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of \u20b9 7 and screws B at a profit of \u20b9 10. Assuming that he can sell all the screws he manufactures, how many pack\u00acages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
\nSolution:
\nLet x be the number of packages of screw A and y be that of screw B.
\nWe have the data as
\n\"NCERT
\nAutomatic machine constraint: 4x + 6y \u2264 240
\nHand operated machine constraint: 6x + 3y \u2264 240
\nProfit function : Z = 7x + 10y
\nThe L.P.P is maximise Z = 1x + 10y subject to the constraints 4x + 6y \u2264 240, 6x + 3y \u2264 240, x, y \u2265 0.
\n\"NCERT
\nMaximum value of Z is 410 at B(30, 20).
\nHence the factory should manufacture 30 packages of screw A and 20 packages of screw B to maximise the profit. The maxi-mum profit is \u20b9 410.<\/p>\n

Question 6.
\nA cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding\/cutting machine and a sprayer. It takes 2 hours on grinding\/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding\/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding\/cutting ma-chine for at the most 12 hours. The profit from the sale of a lamp is \u20b9 5 and that from a shade is \u20b9 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
\nSolution:
\nLet the manufacturer produce x lamps and v wooden shades.
\n\"NCERT
\nGrinding\/cutting constraint: 2x + y \u2264 12
\nSprayer constraint: 3x + 2y \u2264 20
\nProfit function : Z = 5x + 3y
\n\u2234 The L.P.P is maximise Z = 5x + 3y subject to the constraints 2x + y \u2264 12, 3x + 2y \u2264 20, x, y \u2265 0.
\n\"NCERT
\nThe maximum value of Z is 32 at B(4, 4). Hence to get maximum profit 4 pedestal lamps and 4 wooden shades are to be manufactured. The maximum profit obtained is \u20b9 32.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nA company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should be company manufacture in order to maximise the profit?
\nSolution:
\nLet x be the number of souvenirs of type A and y be the number of souvenirs of type B to be manufactured.
\nWe have the data as
\n\"NCERT
\nCutting constraint: 5x+ 8y \u2264 200
\nAssembling constraint: 10x + 8y \u2264 240
\nProfit function : Z = 5x + 6y
\nThen the L.P.P is maximise Z = 5x + 6y subject to the constraints 5x + 8y \u2264 200, 10x + 8y \u2264 240, x, y \u2265 0.
\n\"NCERT
\nThe maximum value of Z is 160 at B(8,20).
\nHence for maximum profit 8 souvenirs of type A and 20 souvenirs of type B should be manufactured.
\nThe maximum profit is \u20b9 160.<\/p>\n

Question 8.
\nA merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost \u20b9 25000 and \u20b9 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than \u20b9 70 lakhs and if his profit on the desktop model is \u20b9 4500 and on portable model is \u20b9 5000.
\nSolution:
\nLet the merchant plan to stock x desktop computers and y portable models.
\nWe have the data as
\n\"NCERT
\nDemand constraint: x + y \u2264 250
\nCost constraint:
\n25000 x + 40000 y \u2264 7000000
\ni.e., 5x + 8 \u2264 1400
\nProfit function : Z = 4500x + 5000y
\nThe L.P.P is maximise Z = 4500x + 5000y
\nsubject to the constraints x + y \u2264 250, 5x + 8y \u2264 1400, x, y \u2265 0.
\n\"NCERT
\nMaximum value of Z is 1150000 at B(200,50) Hence 200 desktop models and 50 portable models are to be stored to get maximum profit.
\nThe maximum profit is \u20b9 1150000.<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nA diet is to contain atlest 80 units of vitamin A and 100 units of minerals. Two foods F1<\/sub> and F2<\/sub> are available. Food F1<\/sub> costs \u20b9 4 per unit food and F2<\/sub> costs \u20b9 6 per unit. One unit of food F1<\/sub> contains 3 units of vitamin A and 4 units of minerals. One unit of food F2<\/sub> contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
\nSolution:
\nLet there be x units of food F1<\/sub> and y units of food F2<\/sub>.
\n\"NCERT
\nVitamin A constraint: 3x + 6v \u2265 80
\nMinarals constraint: 4x + 3y \u2265 100
\nCost function is : Z = 4x + 6y
\nThe L.P.P is minimise Z = 4x + 6y subject to the constraints 3x + 6v \u2265 80,
\n4x + 3y \u2265 100, x,y \u22650.
\n\"NCERT
\nFrom the table the minimum value of Z is 104. Since the region is unbounded, 104 may or may not be the minimum value of Z.
\nConsider the inequality 4x + 6y \u2264 104. This half plane has no point common with the feasible region. Hence the minimum value of Z is 104 at B (24, \\(\\frac { 4 }{ 3 }\\))
\n\u2234 Minimum cost for diet is \u20b9 104<\/p>\n

Question 10.
\nThere are two types of fertilisers F1<\/sub> and F2<\/sub>F1<\/sub> consists of 10% nitrogen and 6% phosphoric acid and F2<\/sub> consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that he needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1<\/sub>, costs \u20b9 6\/kg and F, costs \u20b95\/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
\nSolution:
\nLet x kg of fertilizer F1<\/sub> and y kg of fertilizer
\nF2<\/sub> be used.
\nWe have the data as
\n\"NCERT
\nNitrogen constraint: 0.1 x + 0.05 y \u2265 14
\ni.e., 2x + y \u2265 280
\nPhosphoric acid constraint: 0.06x + 0.1 y \u2265 14
\n3x + 5y \u2265 700
\nCost function : Z = 6x +5y
\nThe L.P.P is minimise Z = 6x + 5y subject to the constraints 2x + y \u2265 280 and 3x + 5y \u2265 700, x \u2265 0, y \u2265 0
\n\"NCERT
\nFrom the table the minimum value of Z is 1000. Since the region is unbounded 1000 may or may not be the minimum value of Z.
\nConsider the inequality 6x + 5y \u2264 1000
\nThis half plane has no point common with the feasible region
\n\u2234 Minimum value of Z is 1000 at B( 100,80).
\nHence 100 kg of fertilizer F1<\/sub> and 80 kg of fertilizer F2<\/sub> is used. The minimum cost of these fertilizers is \u20b9 1000.<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nThe corner points of the feasible region de-termined by the following system of linear inequalities:
\n2x + y \u2264 10, x + 3y \u2264 15, x, y \u2265 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where \u0440, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0,5) is
\na. p = q
\nb. p = 2q
\n\u0441. p = 3q
\nd. q – 3p
\nSolution:
\nd. q – 3p
\nSince maximum Z occurs at (3, 4) and (0, 5), we get Z = 3p + 4q and Z = 0 + 5q
\nSince both values are same.
\n\u2234 3p + 4q = 5q Hence 3p = q<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.2 Question 1. Reshma wishes to mix two types of food P and Q in such a way that the …<\/p>\n

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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.2 Question 1. 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