This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Welcome back Rigan — missed you this week. I will solve question 12. Question 12: A co-operative society has to transport 200 churns of milk. A pick-up and a tractor with a trailer are available. The pick-up can ferry 12 churns per trip and the tractor 20 per trip. The tractor can make a maximum of 5 trips. The total number of trips must not exceed 18. A tractor costs Sh 60 per trip and the pick-up Sh 100 per trip. Find the possible number of trips the two vehicles should make to transport the milk and the corresponding costs. Step 1: Define variables and set up inequalities for the number of churns. Let x be the number of trips made by the pick-up. Let y be the number of trips made by the tractor. The pick-up ferries 12 churns per trip, and the tractor ferries 20 churns per trip. A total of 200 churns must be transported. 12x + 20y 200 This inequality can be simplified by dividing by 4: 3x + 5y 50 Step 2: Set up inequalities for the number of trips. The tractor can make a maximum of 5 trips: y 5 The total number of trips must not exceed 18: x + y 18 Also, the number of trips cannot be negative: x 0 y 0 Step 3: Define the objective function for the cost. The tractor costs Sh 60 per trip, and the pick-up costs Sh 100 per trip. The total cost C is: C = 100x + 60y Step 4: Find possible integer solutions for (x, y) that satisfy all inequalities. We need to find integer pairs (x, y) such that: 1. 3x + 5y 50 2. y 5 3. x + y 18 4. x 0, y 0 Since y 5 and y 0, we can test values for y from 0 to 5. • If y = 0: 3x 50 x 16.67 x 17 x + 0 18 x 18 Possible (x, y) pairs: (17, 0), (18, 0). • If y = 1: 3x + 5 50 3x 45 x 15 x + 1 18 x 17 Possible (x, y) pairs: (15, 1), (16, 1), (17, 1). • If y = 2: 3x + 10 50 3x 40 x 13.33 x 14 x + 2 18 x 16 Possible (x, y) pairs: (14, 2), (15, 2), (16, 2). • If y = 3: 3x + 15 50 3x 35 x 11.67 x 12 x + 3 18 x 15 Possible (x, y) pairs: (12, 3), (13, 3), (14, 3), (15, 3). • If y = 4: 3x + 20 50 3x 30 x 10 x + 4 18 x 14 Possible (x, y) pairs: (10, 4), (11, 4), (12, 4), (13, 4), (14, 4). • If y = 5: 3x + 25 50 3x 25 x 8.33 x 9 x + 5 18 x 13 Possible (x, y) pairs: (9, 5), (10, 5), (11, 5), (12, 5), (13, 5). Step 5: Calculate the cost for each possible combination of trips. The possible (x, y) pairs and their corresponding costs C = 100x + 60y are: • (17, 0): C = 100(17) + 60(0) = 1700 • (18, 0): C = 100(18) + 60(0) = 1800 • (15, 1): C = 100(15) + 60(1) = 1500 + 60 = 1560 • (16, 1): C = 100(16) + 60(1) = 1600 + 60 = 1660 • (17, 1): C = 100(17) + 60(1) = 1700 + 60 = 1760 • (14, 2): C = 100(14) + 60(2) = 1400 + 120 = 1520 • (15, 2): C = 100(15) + 60(2) = 1500 + 120 = 1620 • (16, 2): C = 100(16) + 60(2) = 1600 + 120 = 1720 • (12, 3): C = 100(12) + 60(3) = 1200 + 180 = 1380 • (13, 3): C = 100(13) + 60(3) = 1300 + 180 = 1480 • (14, 3): C = 100(14) + 60(3) = 1400 + 180 = 1580 • (15, 3): C = 100(15) + 60(3) = 1500 + 180 = 1680 • (10, 4): C = 100(10) + 60(4) = 1000 + 240 = 1240 • (11, 4): C = 100(11) + 60(4) = 1100 + 240 = 1340 • (12, 4): C = 100(12) + 60(4) = 1200 + 240 = 1440 • (13, 4): C = 100(13) + 60(4) = 1300 + 240 = 1540 • (14, 4): C = 100(14) + 60(4) = 1400 + 240 = 1640 • (9, 5): C = 100(9) + 60(5) = 900 + 300 = 1200 • (10, 5): C = 100(10) + 60(5) = 1000 + 300 = 1300 • (11, 5): C = 100(11) + 60(5) = 1100 + 300 = 1400 • (12, 5): C = 100(12) + 60(5) = 1200 + 300 = 1500 • (13, 5): C = 100(13) + 60(5) = 1300 + 300 = 1600 The possible number of trips for the pick-up (x) and tractor (y) and their corresponding costs are: &(17, 0) with cost Sh 1700 \\ &(18, 0) with cost Sh 1800 \\ &(15, 1) with cost Sh 1560 \\ &(16, 1) with cost Sh 1660 \\ &(17, 1) with cost Sh 1760 \\ &(14, 2) with cost Sh 1520 \\ &(15, 2) with cost Sh 1620 \\ &(16, 2) with cost Sh 1720 \\ &(12, 3) with cost Sh 1380 \\ &(13, 3) with cost Sh 1480 \\ &(14, 3) with cost Sh 1580 \\ &(15, 3) with cost Sh 1680 \\ &(10, 4) with cost Sh 1240 \\ &(11, 4) with cost Sh 1340 \\ &(12, 4) with cost Sh 1440 \\ &(13, 4) with cost Sh 1540 \\ &(14, 4) with cost Sh 1640 \\ &(9, 5) with cost Sh 1200 \\ &(10, 5) with cost Sh 1300 \\ &(11, 5) with cost Sh 1400 \\ &(12, 5) with cost Sh 1500 \\ &(13, 5) with cost Sh 1600 Drop the next question.