1. **State the problem:** We have two refineries with different costs and production capacities. We want to minimize the total operating cost while meeting the oil order demands for high, medium, and low-grade oil. 2. **Define variables:** Let $x$ = number of days Refinery 1 runs. Let $y$ = number of days Refinery 2 runs. 3. **Objective function:** Minimize cost: $$ \text{Cost} = 20000x + 25000y $$ 4. **Constraints:** We must meet the demand for each grade of oil: - High-grade oil: $$ 400x + 300y \geq 25000 $$ - Medium-grade oil: $$ 300x + 400y \geq 27000 $$ - Low-grade oil: $$ 200x + 500y \geq 30000 $$ Also, non-negativity constraints: $$ x \geq 0, \quad y \geq 0 $$ 5. **Solve the system to minimize cost:** We solve the linear program: Minimize $$20000x + 25000y$$ subject to: $$ 400x + 300y \geq 25000 $$ $$ 300x + 400y \geq 27000 $$ $$ 200x + 500y \geq 30000 $$ $$ x,y \geq 0 $$ 6. **Finding feasible intersection points:** Solve equations by equality to find candidate points: Equation 1 and 2: Multiply eq1 by 4 and eq2 by 3: $$1600x + 1200y = 100000 $$ $$900x + 1200y = 81000 $$ Subtract: $$700x = 19000 \implies x = \frac{19000}{700} = 27.14$$ Use in eq1: $$400(27.14) + 300y = 25000 \implies 10857 + 300y = 25000 \implies 300y = 14143 \implies y = 47.14$$ Check third constraint: $$200(27.14) + 500(47.14) = 5428 + 23570 = 28998 < 30000$$ so not feasible. Equation 1 and 3: $$400x + 300y = 25000$$ $$200x + 500y = 30000$$ Multiply eq3 by 2: $$400x + 1000y = 60000$$ Subtract eq1: $$700y = 35000 \implies y = 50$$ Use in eq1: $$400x + 300(50) = 25000 \implies 400x + 15000 = 25000 \implies 400x = 10000 \implies x = 25$$ Check eq2: $$300(25) + 400(50) = 7500 + 20000 = 27500 \geq 27000$$ feasible. Equation 2 and 3: $$300x + 400y = 27000$$ $$200x + 500y = 30000$$ Multiply eq2 by 5 and eq3 by 4: $$1500x + 2000y = 135000$$ $$800x + 2000y = 120000$$ Subtract: $$700x = 15000 \implies x = 21.43$$ Use in eq2: $$300(21.43) + 400y = 27000 \implies 6429 + 400y = 27000 \implies 400y = 20571 \implies y = 51.43$$ Check eq1: $$400(21.43) + 300(51.43) = 8571 + 15429 = 24000 < 25000$$ not feasible. 7. **Candidate feasible corner point:** $(x, y) = (25, 50)$ 8. **Check other bounds (e.g., $x=0$, $y$ large enough):** At $x=0$, check constraints: $$300y \geq 25000 \Rightarrow y \geq 83.33$$ $$400y \geq 27000 \Rightarrow y \geq 67.5$$ $$500y \geq 30000 \Rightarrow y \geq 60$$ So $y \geq 83.33$ to meet all. Cost: $$20000(0) + 25000(83.33) = 2083333$$ At $y=0$: $$400x \geq 25000 \implies x \geq 62.5$$ $$300x \geq 27000 \implies x \geq 90$$ $$200x \geq 30000 \implies x \geq 150$$ So $x \geq 150$. Cost: $$20000(150) + 25000(0) = 3000000$$ 9. **Calculate cost at $(25, 50)$:** $$20000(25) + 25000(50) = 500000 + 1250000 = 1750000$$ 10. **Conclusion:** Best solution is to run Refinery 1 for 25 days and Refinery 2 for 50 days to minimize cost while meeting demands. **Final answer:** Run Refinery 1 for 25 days and Refinery 2 for 50 days with minimum total cost of 1750000.