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13 NON-FOUNDATION Linear Inequalities in Two Unknowns Name : 13B Date : Mark : 13.2 Linear Programming ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ In each of the following figures, (1 – 2) (a) write down the system of inequalities that determines the shaded region, (b) find the maximum and the minimum values of P = x - y if (x, y) is any point in the shaded region. 1. y 4 x-y=0 2 8x + 3y = 20 -3x + 2y = 5 -3 -2 x -1 0 1 -2 2 3 4 6x + 5y = 4 -4 Solution (a) From the shaded region, take (1, 0) as the test point. For the line 8x + 3y = 20, evaluate the value of 8x + 3y for the point (1, 0). Q 8( 1 \ 8x + 3 y ( ≥ ) + 3( 0 / ) = ( 8 ) ( ≥ / £ ) 20 £ ) 20 is one of the inequalities. For the line 6x + 5y = 4, evaluate the value of 6x + 5y for the point (1, 0). 54 Q 6( 1 \ 6x + 5 y ( ≥ ) + 5( 0 / ) = ( 6 ) ( ≥ / £ )4 £ ) 4 is one of the inequalities. ○ ○ ○ ○ ○ ○ ○ ○ ○ 13 Linear Inequalities in Two Unknowns For the line -3x + 2y = 5, evaluate the value of -3x + 2y for the point (1, 0). Q -3( 1 \ -3x + 2 y ( ≥ \ Ï8 x + 3 y ( ≥ / £ ) 20 Ô The system of inequalities is Ì6 x + 5 y ( ≥ / £ ) 4 . Ô-3x + 2 y ( ≥ / £ ) 5 Ó ) + 2( 0 / ) = ( -3 ) ( ≥ / £ )5 £ ) 5 is one of the inequalities. (b) Draw the line x - y = 0 on the given graph. Translate the line x - y = 0 in the ( positive / negative ) direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at ( 4 , -4 ). \ Maximum value of P = ( 4 ) - ( -4 ) = ( 8 ) Translate the line x - y = 0 in the ( positive / negative ) direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at ( 1 , 4 ). \ Minimum value of P = ( 1 ) - ( 4 ) = ( -3 ) 2. y 2x - 5y = -6 2 1 4x + 7y = 9 x -3 -2 x -y = 0 -1 0 1 -1 2 3 4 x + 7y = -3 Solution (a) From the shaded region, take (0, 0) as the test point. Evaluate the value of 4x + 7y for the point (0, 0). Q 4(0) + 7(0) = 0 £ 9 \ 4x + 7y £ 9 is one of the inequalities. (Solution continues on the next page.) 55 Number and Algebra Evaluate the value of 2x - 5y for the point (0, 0). Q 2(0) - 5(0) = 0 ≥ -6 \ 2x - 5y ≥ -6 is one of the inequalities. Evaluate the value of x + 7y for the point (0, 0). Q (0) + 7(0) = 0 ≥ -3 \ x + 7y ≥ -3 is one of the inequalities. \ Ï4x + 7y £ 9 Ô The system of inequalities is Ì2x - 5y ≥ -6 . Ô Óx + 7y ≥ -3 (b) Draw the line x - y = 0 on the given graph. Translate the line x - y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at ( 4 , -1 ). \ Maximum value of P = 4 - (-1) = 5 Translate the line x - y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at ( -3 , 0 ). \ 56 Minimum value of P = (-3) - 0 = -3 13 Linear Inequalities in Two Unknowns In each of the following, find the maximum and the minimum values of P subject to the given constraints. (3 – 4) 3. P = 2x - 4 y Ïx £ 1 Ô Ì x - 4 y ≥ -7 Ô3x + 2 y ≥ -7 Ó Solution Shade the region that satisfies the following constraints: Ïx £ 1 Ô Ì x - 4 y ≥ -7 Ô3x + 2 y ≥ -7 Ó x - 4 y = -7 3x + 2 y = -7 x -3 -1 1 x -3 -1 1 y 1 1.5 2 y 1 -2 -5 x − 4y = −7 y 2 1 x −3 −2 −1 0 −1 2x − 4y = 0 3x + 2y = −7 1 2 x=1 −2 −3 −4 −5 Draw the line 2 x - 4 y = 0 on the above graph. (Solution continues on the next page.) 57 Number and Algebra Translate the line 2 x - 4 y = 0 in the ( positive / negative ) direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at ( 1 Maximum value of P = 2( \ 1 , -5 ). ) - 4( -5 ) = ( 22 ) Translate the line 2 x - 4 y = 0 in the ( positive / negative ) direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at ( -3 , 1 Minimum value of P = 2( -3 ) - 4( \ 4. ) = ( -10 ) 1 P = x - 3y Ï3x + 5 y ≥ 5 Ô Ì7 x - 5 y ≥ -8 Ô y ≥ 12 x - 62 Ó Solution Shade the region that satisfies the constraints. 3x + 5y = 5 7x - 5y = -8 x -5 0 5 x -4 1 6 y 4 1 -2 y -4 3 10 y = 12x - 62 58 x 5 5.5 6 y -2 4 10 ). 13 Linear Inequalities in Two Unknowns y 12 10 8 7x − 5y = −8 6 y = 12x − 62 4 3x + 5y = 5 2 x − 3y = 0 x −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 −2 −4 Draw the line x - 3y = 0 on the above graph. Translate the line x - 3y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at ( 5 , -2 ). \ Maximum value of P = 5 - 3(-2) = 11 Translate the line x - 3y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at ( 6 , 10 ). \ Minimum value of P = 6 - 3(10) = -24 59 Number and Algebra 5. (a) Shade the region that satisfies all the following constraints: Ï8 x + 6 y £ 31 Ô Ì6 x - 8 y ≥ -33 Ô2 x + 4 y ≥ 9 Ó (b) Find the maximum and the minimum values of P = 3x + 2 y subject to the constraints in (a) if x and y are real numbers, (i) (ii) integers. Solution (a) 8 x + 6 y = 31 6 x - 8 y = -33 x 0.5 2 3.5 x -1.5 0.5 2.5 y 4.5 2.5 0.5 y 3 4.5 6 x -1.5 0.5 2.5 y 3 2 1 2x + 4 y = 9 y 6 6x - 8y = -33 8x + 6y = 31 5 4 3 2 1 3x + 2y = 0 2x + 4y = 9 x -2 60 -1 0 1 2 3 4 13 Linear Inequalities in Two Unknowns Draw the line 3x + 2 y = 0 on the graph in (a). (b) (i) From the graph, P attains its maximum at ( 3.5 , 0.5 ). \ Maximum value of P = 3( 3.5 ) + 2( 0.5 ) = ( 11.5 ) From the graph, P attains its minimum at ( -1.5 , 3 \ Minimum value of P = 3( -1.5 ) + 2( ). ) = ( 1.5 ) 3 (ii) Mark black dots on the same graph to show all the feasible solutions with integral xand y- coordinates. From the graph, P attains its maximum at ( \ Maximum value of P = 3( 3 , 1 3 ) + 2( 1 ) = ( 11 ) From the graph, P attains its minimum at ( -1 , 3 \ Minimum value of P = 3( -1 ) + 2( 6. 3 ). ). ) = ( 3 ) (a) Shade the region that satisfies all the following constraints: Ï6 x + 8 y + 7 £ 0 Ô Ì10 x + 4 y + 21 ≥ 0 Ô2 x + 12 y + 21 ≥ 0 Ó (b) Find the maximum and the minimum values of P = 3x - y + 2 subject to the constraints in (a) if x and y are real numbers, (i) (ii) integers. Solution (a) 6x + 8y + 7 = 0 x y 10x + 4y + 21 = 0 -2.5 -0.5 1.5 x -0.5 -2 y 1 -2.5 -1.5 -0.5 1 -1.5 -4 2x + 12y + 21 = 0 x -1.5 1.5 4.5 y -1.5 -2 -2.5 (Solution continues on the next page.) 61 Number and Algebra y 10x + 4y + 21 = 0 3x − y + 2 = 0 2 1 x −3 −2 −1 0 1 2 3 4 −1 −2 2x + 12y + 21 = 0 −3 −4 (b) (i) 6x + 8y + 7 = 0 Draw the line 3x - y + 2 = 0 on the graph in (a). From the graph, P attains its maximum at ( 1.5 , -2 ). \ Maximum value of P = 3(1.5) - (-2) + 2 = 8.5 From the graph, P attains its minimum at ( -2.5 , 1 ). \ Minimum value of P = 3(-2.5) - 1 + 2 = -6.5 (ii) Mark black dots on the same graph to show all the feasible solutions with integral xand y- coordinates. From the graph, P attains its maximum at ( 0 , -1 ). \ Maximum value of P = 3(0) - (-1) + 2 = 3 From the graph, P attains its minimum at ( -2 , 0 ). \ Minimum value of P = 3(-2) - 0 + 2 = -4 62