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II P U C MATHEMATICS, Sree Guru, Amma, , LINEAR PROGRAMMING, PROBLEMS, 1-M+6-M=7 Marks, , ONE MARK QUESTIONS, Ex: Define feasible region., Solution: A set of values of the variables, satisfying all the constraints is known as a, feasible region., Ex: In Linear programming problems,, define linear objective function., Solution: Linear function z  ax  by ,, where a, b are constant, which has to be, maximised or minimized is called a linear, objective function., Ex: Define optimal solution in a linear, programing problem., Solution: Any point in the feasible region, , April-May - 2022, , SIX MARK QUESTIONS, Ex: Solve the following problem, graphically: Maximize and minimize, z  3x  2 y subject to constraints x  2 y  10,, 3x  y  15, x, y  0 ., Solution: Region represented by x  2 y  10, Consider x  2 y  10 …(1), , , , x y,  1, 10 5, ,  (1) passes through A (10, 0) and, B (0, 5). Join AB and produce it both, ways., Put x  0, y  0  0  0  10 which is true.,  Solution region contains the origin., Region represented by 3x  y  15, Consider 3x  y  15 …(2), , , , x y,  1, 5 15, ,  (2) passes through C (5, 0) and, D (0,15).join CD and produce it both ways., Put x  0, y  0  0  0  15 which is true.,  Solution region contains the origin., Solve (1) and (2) x  4, y  3  E (4, 3), , that gives the optimal value (maximum or, minimum) of the objective function is, called an optimal solution linear, programing problem., “OR”, The maximum or minimum values of the, objective function are called optimal, solution., Ex: Define Constraints in Linear, programming problems., Solution: The set of points satisfying the, constraints of an LPP is said to be, constrained set of LPP., , Solution lies in the region OBEC., Sl., No., , Corner, points, , Corresponding, value of a, z  3x  2 y, , O (0, 0), O, B (0, 5), 10 Minimum, E (4, 3), 18  Maximum, C (5, 0), 15, Hence, maximum value of z is 18 at the, point (4, 3) and Minimum value of z is 10, at the point (0, 5)., 1., 2., 3., 4., , Bhadrasetty-9901782340, sujm pu collge, Harapanahalli – 583131., , Page 1

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II P U C MATHEMATICS, Ex: Miximise z  4 x  y subject to, Constraints x  y  50, 3x  y  90,, x  0, y  0 by graphical method., Solution: Region represented by x  y  50 , Consider x  y  50 …(1), Put x  0  y  50, ,  A(0, 50), , Put y  0  x  50, ,  B(50,0), , Plot the points A (0, 50) and B (50, 0). Join, AB and produce it both ways., Clearly O (0, 0) satisfies x  y  50 ,  Solution region is on origin side., Region represented by 3x  y  90, Consider 3x  y  90 …(2), If x  0 then y  90  C (0,90), If y  0 then x  30  D(30,0), Put x  0, y  0 then 0  0  90 which is true., Therefore solution region contains the, origin., Solve (1) and (2), we get x  20, y  30, , April-May - 2022, Ex: Minimize z  3x  4 y subject to, the constraints x  2 y  8 3x  2 y  12, x  0, y  0 , by graphical method., , Solution: Region represented by x  2 y  8, Consider x  2 y  8, , …(1), , Put x  0  y  4, ,  A (0, 4), , Put y  0  x  8, ,  B (8, 0), , Plot the points A (0, 4) and B (8, 0). Join, AB and produce it both ways., Put x  0, y  0  0  0  8 which is true.,  Solution region contains the origin., Region represented by 3x  2 y  12, Consider 3x  2 y  12, , ..…(2), , Put x  0  y  6, ,  C (0, 6), , Put y  0  x  4, ,  D (4, 0), , Plot the points C (0, 6) and D (4, 0). Join, CD and produce it both ways., Put x  0, y  0  0  0  12 which is true.,  Solution region contains the origin., Solve (1) and (2) we get x  2 and y  3, ,  E(20,30), , E (2, 3)., ,  Solution lies in the region OAED., Corresponding, Sl., Corner points, value of z  4 x  y, No., , Solution lies in the region OAED., Corresponding, Sl., Corner, value of z  3x  4 y, No., points, , 1., , O (0, 0), , 0, , 1., , O (0, 0), , 0, , 2., , A (0, 50), , 50, , 2., , A (0, 4), , 16, , 3., , E (20, 30), , 110, , 3., , E (2, 3), , 6, , D (30, 0), , 120  Maximum, , 4., , D (4, 0), ,  12  Minimum, , 4., , Hence, maximum value of Z is 120 at the, point (30, 0)., , Hence, minimum value of z is  12 at the, point (4, 0)., , Bhadrasetty-9901782340, sujm pu collge, Harapanahalli – 583131., , Page 2

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II P U C MATHEMATICS, Ex: Solve the following problem, , April-May - 2022, Ex: Minimize and Maximize z  x  2 y, , graphically: Maximize and minimize, , subject to the constraints x  2 y  100,, , z  10500 x  9000 y subject to constraints, x  y  50 , 2 x  y  80 , x  0 , y  0 ., , Solution: From the given data,, z  10500 x  9000 y subject to the, constraints: x  y  50 , 2 x  y  80 x, y  0, Region represented by x  y  50, Consider x  y  50, …(1), It passes through A (50, 0) and B (0,, 50). Join AB and produce it both ways., Put x  0, y  0  0  0  50 is true.,  Solution region constrains, origin., Region represented by 2 x  y  80, Consider 2 x  y  80, …(2), , , the, , x, y,  1, 40 80, ,  It passes through C (40, 0) and D (0,, 80). Join CD and produce it both ways., Put x  0, y  0  0  0  80 is true.,  Solution region contains the origin., Solve (1) and (2) we get x  30 and y  20,  E (30, 20), , 2 x  y  0,, , 2 x  y  200, x, y  0 by the, , graphical method., Solution: Minimize and Maximize,, z  x  2y, , Draw the graph of the line x  2 y  100, If x  0 when y  50 and x  100 when y  0, Putting (0, 0) in the inequality x  2 y  100 ,, we have, 0  100 is false, so the half plane, is away from the origin., Secondly, draw the graph of the line, 2x  y  0, , Then x  0; y  0 and x  10; y  20, Putting (1, 1) in the inequality 2 x  y  0 ,, we have 1  0 is false, so the half plane is, towards y  axis ., Thirdly, draw the graph of the line, 2 x  y  200, , Then x  0; y  200 and x  100; y  0, Putting (0, 0) in the inequality 2 x  y  200 ,, we have 0  200 is true, so the half plane,  Solution region is OBEC., Value of, Sl., Corner, z  10500 x  9000 y, No., point, O (0, 0), 0, 1., B (0, 50) 450000, 2., E (30, 20) 495000  Maximum, 3., C (40, 0) 420000 Minimum, 4., ∴Minimum value of Z=420000 at the point, C (40, 0), Maximum value of Z=495000 at the point, E (30, 20)., , is towards the origin., Since x, y  0 , so the feasible region lies in, the first quadrant., On solving equations x  2 y  100 and, 2 x  y  0 , we get, (20, 40), , On solving equations 2 x  y  0 and, 2 x  y  200 , we get, (50,100) ., , Bhadrasetty-9901782340, sujm pu collge, Harapanahalli – 583131., , Page 3

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II P U C MATHEMATICS, The corner points of the feasible region are, ,  0,50 ,  20, 40 , 50,100 and  0, 200 ., , Type, , No., of, cakes, , April-May - 2022, Flour, Fat, required in, required in, gms, gms, , I, , x, , 200 x, , 25 x, , II, , y, , 100 y, , 50 y, , 200 x + 100 y, , 25 x + 50 y, , Total x + y, ,  Total number of cakes z  x  y, such that 200 x  10 y  5000 (flour, constraints), The values of Z at these points are as, follows:, Corner point, , z  x  2y, , (0,50), , 100  Minimum, , (20, 40), , 100  Minimum, , (50,100), , 250, , (0, 200), , 400  Maximum, , The maximum value of Z is 400 at, (0, 200) and the minimum value of Z, , is 100 at all the points on the line, segment joining (0,50) and (20, 40) ., Ex: One kind of cake requires 200 g of, flour and 25 g of fat, and another kind of, cake requires 100 g flour and 50 g of fat., , 25 x  50  1000, , (fat constraints), , Hence mathematically, we have, maximize z  x  y subject to constraints, 2 x  y  50, x  2 y  40, x, y  0, , Region represented by 2 x  y  50, Consider 2 x  y  50 …(1) , , x, y, , 1, 25 50, ,  It passes through A (25, 0) and B (0,, 50). Join AB and produce it both ways., Put x  0, y  0  0  0  50 is true.,  Solution region contains the origin., Region represented by x  2 y  40, Consider x  2 y  40, , …(2), , x, y, , 1, 40 20, , Find the maximum number of cakes, , , , which can be made from 5 kg flour and 1, , It passes through C (40, 0) and D (0, 20)., , kg of fat assuming that there is no, , Join CD and produce it both ways., , shortage of the other ingredients, used in, making the cakes., , Put x  0, y  0  0  0  40 is true.,  Solution region contains the origin., , Solution: Let the number of cakes made, , Solve (1) and (2) we get x  20, y  10, , of type I be x and that of type II be y., ,  E (20, 10) , , Bhadrasetty-9901782340, sujm pu collge, Harapanahalli – 583131., , Page 4

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II P U C MATHEMATICS, , April-May - 2022, Consider x  y  60 …(2) , , x, y, , 1, 60 60, ,  (2) passes through C (60, 0) and, B (0, 60). Join CB and produce it both, ways., Put x  y  0  0  0  60 which is false., ,  Solution region is ODEA., Sl., No., , Corner point, ,  Solution region does not contain the, , Value of z  x  y, , origin., Region represented by x  2 y  0, , 1., , O (0, 0), , 0, , 2., , D (0, 20), , 20, , It passes through the origin and, , 3., , E (20, 10), , 30  Maximum, , E (60, 30). Join OE and produce it both, , 4., , A (25, 0), , Consider x  2 y  0, , …(3), , ways., 25, , Put x  30 and y = 0  30  0  0 which is, , Hence, the maximum value of z is 30 at, , true., , E (20, 10)., , Solution region contains point (30, 0)., , Ex: Minimize and maximize z  5x  10 y, subject to constraints x  2 y  120, x  y  60,, x  2 y  0, x, y  0 ., , Solution: Region represented by, x  2 y  120 , , Consider x  2 y  120 …(1) , , x, y, , 1, 120 60, , Solution lies in the region CEDA., Value of, ,  (1) passes through A (120, 0) and, , Sl., No., , B (0, 60). Join AB and produce it both, , 1., , C (60, 0), , 300  Minimum, , 2., , E (40, 20), , 400, , 3., , D (60, 30), , 600 Maximum, , 4., , A (120, 0), , 600 Maximum, , ways., Put x  0, y  0  0  0  120 which is true.,  Solution region contains the origin., Region represented by x  y  60, , Corner, point, , z  5x  10 y, , Bhadrasetty-9901782340, sujm pu collge, Harapanahalli – 583131., , Page 5

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II P U C MATHEMATICS, Hence, minimum value of z is 300 at (60,, , April-May - 2022, If x = 1 then y = 1  E (1, 1), , 0) and maximum value of z is 600 at all, , Plot the points F (2, 2) and E (1, 1). Join, , the points on the line segment joining the, , EF and produce it both ways., , points A (120, 0) and D (60, 30)., , Clearly O (0, 1) satisfies x  y  0   (0, 1), lies in the region x  y  0 , , Ex: Minimize and maximize Z  3x  9 y,, , On solving x  y and x  y  10, we get the, , subject to the constrains x  3 y  60;, , point G (5, 5), , x  y  10; x  y; x, y  0 ., , Solution: Region represented by x  3 y  60, , On solving x  y and x  3 y  60, we get the, point H (15, 15)., , Consider the equation x  3 y  60, Put x  0  y  20  A(0,20), Put y  0  x  60  B(60,0), Plot the points A (0, 20) and B (60,0). Join, AB and produce it both ways., Putting x = 0 and y = 0, we get 0  3(0)  60 ,, is true.,  O (0, 0) lies in the region x  3 y  60  So,, , Thus, the feasible region is ACGH, as, , the region containing the origin is the, , shown in the figure., , solution set of x  3 y  60 ., , Sl., , Corner, , Corresponding, , Region represented by x  y  10 ., , No., , points, , value of z  3 x  9 y, , Consider the equation x  y  10 ., , 1., , C (0, 10), , 90, , Put x  0  y  10  C (0,10), , 2., , G (5, 5), , 60  Minimum, , Put y  0  x  10  D(10,0), , 3., , H (15, 15), , 180 Maximum, , 4., , A (0, 20), , 180 Maximum, , Plot the points C (0, 10) and D (10, 0)., Join CD and produce it both ways., Now, x  0, y  0  0  0  10 , is not true.,  O (0, 0) does not lie in the region, x  y  10 ., ,  The minimum value of Z is 60, and, maximum value of Z is 180., @@@@@@@@@, , Region represented by x  y i.e., x  y  0, Consider x  y i.e., x  y  0, If x  2 then y  2  F (2, 2), Bhadrasetty-9901782340, sujm pu collge, Harapanahalli – 583131., , Page 6