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.
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Step 1: Define variables and formulate inequalities (Part a) Let x be the number of kg of Type X feed. Let y be the number of kg of Type Y feed. The constraints are based on the minimum vitamin requirements: • Vitamin A requirement: The cow needs at least 160 units of Vitamin A. Type X provides 9 units/kg, and Type Y provides 20 units/kg. 9x + 20y 160 • Vitamin B requirement: The cow needs not less than 160 units of Vitamin B. Type X provides 4 units/kg, and Type Y provides 40 units/kg. 4x + 40y 160 This inequality can be simplified by dividing by 4: x + 10y 40 • Non-negativity constraints: The amount of feed cannot be negative. x 0 y 0 The objective function to minimize is the total cost C: • Cost: Type X feed costs 80 units/kg, and Type Y feed costs 100 units/kg. C = 80x + 100y The inequalities are: 9x + 20y 160 x + 10y 40 x 0 y 0 Step 2: Graph the inequalities and show the feasible region (Part b) To graph the inequalities, we first consider the boundary lines: 1. L_1: 9x + 20y = 160 • If x = 0, 20y = 160 y = 8. Point: (0, 8) • If y = 0, 9x = 160 x = (160)/(9) ≈ 17.78. Point: ((160)/(9), 0) The region 9x + 20y 160 is above or to the right of this line. 2. L_2: x + 10y = 40 • If x = 0, 10y = 40 y = 4. Point: (0, 4) • If y = 0, x = 40. Point: (40, 0) The region x + 10y 40 is above or to the right of this line. The feasible region is the area in the first quadrant (x 0, y 0) that satisfies both 9x + 20y 160 and x + 10y 40. This region is unbounded. To find the corner points of the feasible region, we find the intersection of the boundary lines: • Intersection of L_1 and L_2: From x + 10y = 40, we have x = 40 - 10y. Substitute into 9x + 20y = 160: 9(40 - 10y) + 20y = 160 360 - 90y + 20y = 160 360 - 70y = 160 70y = 200 y = (200)/(70) = (20)/(7) Now find x: x = 40 - 10((20)/(7)) = 40 - (200)/(7) = (280 - 200)/(7) = (80)/(7) Intersection point: ((80)/(7), (20)/(7)) The corner points of the feasible region are: • (0, 8) (intersection of L_1 and the y-axis) • ((80)/(7), (20)/(7)) (intersection of L_1 and L_2) • (40, 0) (intersection of L_2 and the x-axis) Graph description: Draw an x-axis and a y-axis. Plot the line 9x + 20y = 160 by connecting (0, 8) and ((160)/(9), 0) ≈ (17.78, 0). Plot the line x + 10y = 40 by connecting (0, 4) and (40, 0). The feasible region is the area in the first quadrant (x 0, y 0) that lies above both lines. This region is bounded by the y-axis from y=8 upwards, by the line 9x+20y=160 from (0,8) to ((80)/(7), (20)/(7)), by the line x+10y=40 from ((80)/(7), (20)/(7)) to (40,0), and by the x-axis from x=40 rightwards. Step 3: Determine the least cost (Part c) To find the least cost, we evaluate the objective function C = 80x + 100y at each corner point of the feasible region: • At (0, 8): C = 80(0) + 100(8) = 0 + 800 = 800 • At ((80)/(7), (20)/(7)): C = 80((80)/(7)) + 100((20)/(7)) = (6400)/(7) + (2000)/(7) = (8400)/(7) = 1200 • At (40, 0): C = 80(40) + 100(0) = 3200 + 0 = 3200 Comparing the costs, the minimum cost is 800. The least cost of maintaining a cow at Jimmy's farm is 800 units, achieved by using 0 kg of Type X feed and 8 kg of Type Y feed. a) The inequalities are: 9x + 20y 160 x + 10y 40 x 0 y 0 b) The feasible region is the area in the first quadrant bounded by the lines 9x + 20y = 160 and x + 10y = 40, lying above both lines. The corner points of this region are