1. **Problem Statement:** A firm operates two plants, A and B, producing cakes, pellets, and meal. We want to find how long each plant should run to meet orders of 2000 tonnes of cakes, 500 tonnes of pellets, and 1350 tonnes of meal, minimizing cost. 2. **Define variables:** Let $x$ = hours Plant A runs, $y$ = hours Plant B runs. 3. **Constraints from production:** - Cakes: $25x + 50y \geq 2000$ - Pellets: $15x + 5y \geq 500$ - Meal: $10x + 75y \geq 1350$ 4. **Objective function (cost to minimize):** $$C = 200x + 250y$$ 5. **Solve constraints as equalities to find intersection points:** - From cakes and pellets: $$25x + 50y = 2000$$ $$15x + 5y = 500$$ Multiply second by 10: $$150x + 50y = 5000$$ Subtract first: $$(150x + 50y) - (25x + 50y) = 5000 - 2000 \Rightarrow 125x = 3000 \Rightarrow x = 24$$ Substitute $x=24$ into cakes: $$25(24) + 50y = 2000 \Rightarrow 600 + 50y = 2000 \Rightarrow 50y = 1400 \Rightarrow y = 28$$ - Check meal constraint: $$10(24) + 75(28) = 240 + 2100 = 2340 \geq 1350$$ (satisfied) 6. **Check if this point meets all constraints:** Yes. 7. **Calculate cost:** $$C = 200(24) + 250(28) = 4800 + 7000 = 11800$$ 8. **Check other intersections for lower cost:** - Cakes and meal: $$25x + 50y = 2000$$ $$10x + 75y = 1350$$ Multiply first by 3: $$75x + 150y = 6000$$ Multiply second by 2.5: $$25x + 187.5y = 3375$$ Subtract second from first: $$(75x + 150y) - (25x + 187.5y) = 6000 - 3375 \Rightarrow 50x - 37.5y = 2625$$ Solve for $x$ or $y$ is complex; alternatively, solve system: From cakes: $$25x + 50y = 2000 \Rightarrow x = \frac{2000 - 50y}{25} = 80 - 2y$$ Substitute into meal: $$10(80 - 2y) + 75y = 1350 \Rightarrow 800 - 20y + 75y = 1350 \Rightarrow 55y = 550 \Rightarrow y = 10$$ Then $x = 80 - 2(10) = 60$ Check pellets: $$15(60) + 5(10) = 900 + 50 = 950 \geq 500$$ (satisfied) Cost: $$200(60) + 250(10) = 12000 + 2500 = 14500$$ (more expensive) 9. **From pellets and meal:** $$15x + 5y = 500$$ $$10x + 75y = 1350$$ Multiply first by 15: $$225x + 75y = 7500$$ Subtract second: $$(225x + 75y) - (10x + 75y) = 7500 - 1350 \Rightarrow 215x = 6150 \Rightarrow x = 28.6$$ Substitute $x$ into pellets: $$15(28.6) + 5y = 500 \Rightarrow 429 + 5y = 500 \Rightarrow 5y = 71 \Rightarrow y = 14.2$$ Check cakes: $$25(28.6) + 50(14.2) = 715 + 710 = 1425 \not\geq 2000$$ (not satisfied) 10. **Conclusion:** The optimal running times are $x=24$ hours for Plant A and $y=28$ hours for Plant B with minimum cost 11800. **Final answer:** $$\boxed{x=24\text{ hours},\quad y=28\text{ hours},\quad \text{Minimum cost} = 11800}$$