B40000 104 Vectors and Three Dimensional Geometry 4 A tailor needs at least 40

B40000 104 vectors and three dimensional geometry 4 a

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104 Vectors and Three Dimensional Geometry 4. A tailor needs at least 40 large buttons and 60 small buttons. In the market, buttons are available in boxes or cards. A box contains 6 large and two small buttons and a card contains 2 large and 4 small buttons. If the cost of a box is 30 paise and card is 20 paise. Find how many boxes and cards should he buy as to minimise the expenditure ? 4.5 ANSWERS TO CHECK YOUR PROGRESS 1. Type A : x , Type B : y then LPP is Maximize P = 50 x + 25 y subject to 5 x + 8 y ≤ 200 [Cutting constraint] 10 x + 8 y ≤ 240 [Assembly constraint] x ≥ 0 , y ≥ 0 [Non- negativity] P(A) = 1200 P (B) = 900 P (C) = 625 P(0) = 0 Thus, profit is maximum when x = 24, y = 0 Maximum profit = Rs. 1200 2. Chairs : x , Tables : y , then LPP is Maximize P = 20 x + 30 y subject to 3 x + 3 y ≤ 36 [Machine A 1 constraint] 5 x + 2 y ≤ 50 [Machine A 2 constraint] 2 x + 6 y ≤ 60 [Machine A 3 constraint] x ≥ 0, y 0 [Non-negativity] Y 30 25 20 10 0 10 20 A 30 40 x C B(8,20) Figure 19
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105 Linear Programming 30 20 10 0 10 20 30 x Y 3x + 3y = 36 5x + 2y = 50 2x + 6y = 60 D C B A P (A) = 20 (10) + 30(0) = 200 26 10 820 (B) = 20 + 30 = 3 3 3 P P (C) = 20 (3) + 30(9) = 330 P (D) = 20 (10) + 30(0) = 200 P(0) = 20(0) + 30(0) = 0 Thus, profit is maximum when x = 3, y = 9. Maximum Profit = Rs. 330 3. First type trunks : x Second type trunks : y The LPP is Maximise P = 30 x + 40 y , subject to 3 x + 3 y ≤ 18 [Machine A constraint] 2 x + 3 y ≤ 14 [Machine B constraint] x ≥ 0, y ≥0 [non-negativity] P(A) = 180 P(B) =200 P (C) = 560/3 P(O) = 0 Figure 20
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106 Vectors and Three Dimensional Geometry Thus, profit is maximum when x = 4, y = 2 and maximum profit is Rs. 200. 4. We write the information in the question in tabular form as follows : First type of machine ( x ) First type of machine ( x ) Constraint Area 20 (sq. m per machine) 24 (sq. m per machine) 200 Labour 5 (per machine) 3 (per machine) ≤ 40 Buckets 120 (per machine) 80 (per machine) Maximise N Let x machines of the first type and y machines of the second type be purchased. We have to Maxmise N = 120 x + 80 y subject to 20 x + 24 y ≤ 200 (Area constraint) 5 x + 3 y ≤ 40 (Labour constraint) x ≥ 0, y ≥ 0 ( Non-negativity) 4 6 C 3 B(4,2) 0 3 6 A 14/3 3 6 x Figure 21
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107 Linear Programming Y 13 25/3 8 6 4 2 0 2 4 6 8 10 x A(0,25/3) B(6,10/3) 5 x + 3 y = 4 0 20 x + 2 4 y = 2 0 0 C(8,0) Figure 22 The feasible region for the above linear programming problem has been shaded in the figure. We find the value of N at the cornor points of the feasible region. We have 25 2,000 2 0, = = 666 3 3 3 N (A) = N 10 2960 2 N (B) = N 6, = = 986 3 3 3 N (C) = N(8,0) = 960 N (O) = N(0,0) = 0 Thus, the value of N is maximum when x = 6, y = 10/3 . As y cannot be in fraction, we take x = 6, y = 3. 5. We summarise the information given in the question in tabular form as follows: Good X Good Y Constraint Capital 2 1 10 Labour 1 3 20 Revenue 80 100 Maximize R Let x units of X and y units of Y be produced. The above problem can be written as Maxmize R = 80 x + 100 y subject to 2 x + y ≤ 10 (Capital constraint) x + 3 y ≤ 20 (Labour constraint) x ≥ 0, y ≥ 0 ( Non-negativity) We shade the feasible region in following figure.
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