Question:

Step 1: Define Variables Let $x$ be the number of days Factory I operates. Let $y$ be the number of days Factory II operates. Step 2: Formulate the Objective Function The objective is to minimize the total operating cost $Z$. The cost for Factory I is ₹12000 per day, and for Factory II is ₹15000 per day. $$Z = 12000x + 15000y$$ Step 3: Establish Constraints The company must meet the minimum demand for each model. For Model A: 50 calculators from Factory I and 40 from Factory II per day. Demand is at least 6400. $$50x + 40y \ge 6400 \implies 5x + 4y \ge 640$$ For Model B: 50 calculators from Factory I and 20 from Factory II per day. Demand is at least 4000. $$50x + 20y \ge 4000 \implies 5x + 2y \ge 400$$ For Model C: 30 calculators from Factory I and 40 from Factory II per day. Demand is at least 4800. $$30x + 40y \ge 4800 \implies 3x + 4y \ge 480$$ Additionally, the number of days cannot be negative: $$x \ge 0$$ $$y \ge 0$$ Step 4: Determine the Feasible Region and its Corner Points The feasible region is defined by the intersection of these inequalities. The corner points of this region are found by solving pairs of the corresponding equality equations and verifying feasibility. 1. Intersection of $5x + 4y = 640$ and $5x + 2y = 400$: Subtracting the second equation from the first: $(5x + 4y) - (5x + 2y) = 640 - 400$ $2y = 240 \implies y = 120$ Substitute $y=120$ into $5x + 2y = 400$: $5x + 2(120) = 400 \implies 5x + 240 = 400 \implies 5x = 160 \implies x = 32$ This gives the point $(32, 120)$. Verification: Model A: $50(32) + 40(120) = 1600 + 4800 = 6400 \ge 6400$ Model B: $50(32) + 20(120) = 1600 + 2400 = 4000 \ge 4000$ Model C: $30(32) + 40(120) = 960 + 4800 = 5760 \ge 4800$ The point $(32, 120)$ is feasible. 2. Intersection of $5x + 4y = 640$ and $3x + 4y = 480$: Subtracting the second equation from the first: $(5x + 4y) - (3x + 4y) = 640 - 480$ $2x = 160 \implies x = 80$ Substitute $x=80$ into $3x + 4y = 480$: $3(80) + 4y = 480 \implies 240 + 4y = 480 \implies 4y = 240 \implies y = 60$ This gives the point $(80, 60)$. Verification: Model A: $50(80) + 40(60) = 4000 + 2400 = 6400 \ge 6400$ Model B: $50(80) + 20(60) = 4000 + 1200 = 5200 \ge 4000$ Model C: $30(80) + 40(60) = 2400 + 2400 = 4800 \ge 4800$ The point $(80, 60)$ is feasible. 3. Intersection with the axes: - For $x=0$: $40y \ge 6400 \implies y \ge 160$ $20y \ge 4000 \implies y \ge 200$ $40y \ge 4800 \implies y \ge 120$ To satisfy all constraints, $y$ must be at least 200. Thus, $(0, 200)$ is a corner point. Verification: Model A: $50(0) + 40(200) = 8000 \ge 6400$ Model B: $50(0) + 20(200) = 4000 \ge 4000$ Model C: $30(0) + 40(200) = 8000 \ge 4800$ The point $(0, 200)$ is feasible. - For $y=0$: $50x \ge 6400 \implies x \ge 128$ $50x \ge 4000 \implies x \ge 80$ $30x \ge 4800 \implies x \ge 160$ To satisfy all constraints, $x$ must be at least 160. Thus, $(160, 0)$ is a corner point. Verification: Model A: $50(160) + 40(0) = 8000 \ge 6400$ Model B: $50(160) + 20(0) = 8000 \ge 4000$ Model C: $30(160) + 40(0) = 4800 \ge 4800$ The point $(160, 0)$ is feasible. The feasible corner points are $(0, 200)$, $(160, 0)$, $(32, 120)$, and $(80, 60)$. Step 5: Evaluate the Objective Function at Each Corner Point Substitute the coordinates of each feasible corner point into the objective function $Z = 12000x + 15000y$: - At $(0, 200)$: $Z = 12000(0) + 15000(200) = 0 + 3,000,000 = 3,000,000$ - At $(160, 0)$: $Z = 12000(160) + 15000(0) = 1,920,000 + 0 = 1,920,000$ - At $(32, 120)$: $Z = 12000(32) + 15000(120) = 384,000 + 1,800,000 = 2,184,000$ - At $(80, 60)$: $Z = 12000(80) + 15000(60) = 960,000 + 900,000 = 1,860,000$ Step 6: Determine the Minimum Cost Comparing the costs at the feasible corner points, the minimum operating cost is $₹1,860,000$. This occurs when Factory I operates for 80 days and Factory II operates for 60 days.