1. **Problem Statement:** We have two sets of cost functions for four communities each, representing total cost as a function of quantity $x$. We need to plot cost curves, find break-even points, and determine the best location for given demands. --- ### For Mr. Irfan's communities: 2. **Cost functions:** - Dhaka: $y = 250000 + 60x$ - Chittagong: $y = 400000 + 35x$ - Khulna: $y = 700000 + 20x$ - Barisal: $y = 600000 + 30x$ 3. **Break-even analysis:** Set pairs equal to find intersection points (break-even quantities): - Dhaka vs Chittagong: $$250000 + 60x = 400000 + 35x \Rightarrow 25x = 150000 \Rightarrow x = 6000$$ - Chittagong vs Barisal: $$400000 + 35x = 600000 + 30x \Rightarrow 5x = 200000 \Rightarrow x = 40000$$ - Barisal vs Khulna: $$600000 + 30x = 700000 + 20x \Rightarrow 10x = 100000 \Rightarrow x = 10000$$ - Chittagong vs Khulna: $$400000 + 35x = 700000 + 20x \Rightarrow 15x = 300000 \Rightarrow x = 20000$$ - Dhaka vs Barisal: $$250000 + 60x = 600000 + 30x \Rightarrow 30x = 350000 \Rightarrow x = 11666.67$$ - Dhaka vs Khulna: $$250000 + 60x = 700000 + 20x \Rightarrow 40x = 450000 \Rightarrow x = 11250$$ 4. **Ranges where each community is cheapest:** - For low $x$, Dhaka is cheapest due to low fixed cost despite high variable cost. - Between $x=6000$ and $x=10000$, Barisal becomes cheaper than Dhaka and Chittagong. - For $x > 10000$, Khulna is cheapest due to lowest variable cost. 5. **Best location for demand 14,000 units:** Calculate total costs: - Dhaka: $250000 + 60 \times 14000 = 250000 + 840000 = 1,090,000$ - Chittagong: $400000 + 35 \times 14000 = 400000 + 490000 = 890,000$ - Khulna: $700000 + 20 \times 14000 = 700000 + 280000 = 980,000$ - Barisal: $600000 + 30 \times 14000 = 600000 + 420000 = 1,020,000$ **Best location:** Chittagong with cost 890,000. --- ### For Mr. Robin's communities: 6. **Cost functions:** - Dhaka: $y = 250000 + 11x$ - Chittagong: $y = 100000 + 30x$ - Khulna: $y = 150000 + 20x$ - Barisal: $y = 200000 + 35x$ 7. **Break-even analysis:** - Dhaka vs Chittagong: $$250000 + 11x = 100000 + 30x \Rightarrow 19x = 150000 \Rightarrow x = 7894.74$$ - Dhaka vs Khulna: $$250000 + 11x = 150000 + 20x \Rightarrow 9x = 100000 \Rightarrow x = 11111.11$$ - Dhaka vs Barisal: $$250000 + 11x = 200000 + 35x \Rightarrow 24x = 50000 \Rightarrow x = 2083.33$$ - Chittagong vs Khulna: $$100000 + 30x = 150000 + 20x \Rightarrow 10x = 50000 \Rightarrow x = 5000$$ - Chittagong vs Barisal: $$100000 + 30x = 200000 + 35x \Rightarrow 5x = -100000 \Rightarrow x = -20000$$ (no positive intersection) - Khulna vs Barisal: $$150000 + 20x = 200000 + 35x \Rightarrow 15x = -50000 \Rightarrow x = -3333.33$$ (no positive intersection) 8. **Ranges where each community is cheapest:** - For $x < 2083$, Barisal is cheapest. - Between $2083$ and $5000$, Chittagong is cheapest. - Between $5000$ and $7894.74$, Khulna is cheapest. - For $x > 7894.74$, Dhaka is cheapest. 9. **Best location for demand 8,000 units:** Calculate total costs: - Dhaka: $250000 + 11 \times 8000 = 250000 + 88000 = 338,000$ - Chittagong: $100000 + 30 \times 8000 = 100000 + 240000 = 340,000$ - Khulna: $150000 + 20 \times 8000 = 150000 + 160000 = 310,000$ - Barisal: $200000 + 35 \times 8000 = 200000 + 280000 = 480,000$ **Best location:** Khulna with cost 310,000. --- **Summary:** - Mr. Irfan's best location for 14,000 units is Chittagong. - Mr. Robin's best location for 8,000 units is Khulna.