1. **State the problem:** We want to minimize the operating cost of two refineries meeting specific oil production demands. 2. **Define variables:** Let $x$ = number of days Refinery 1 runs Let $y$ = number of days Refinery 2 runs ### a) Objective function: Minimize cost: $$ C = 20000x + 25000y $$ ### b) Constraints: From production: - High-grade oil: $$400x + 300y \geq 25000$$ - Medium-grade oil: $$300x + 400y \geq 27000$$ - Low-grade oil: $$200x + 500y \geq 30000$$ Also, days must be non-negative: $$x \geq 0,\quad y \geq 0$$ ### c) Solve for $x$ and $y$ to minimize costs: Use constraints as equalities to find intersection points: Solve system: $$\begin{cases} 400x + 300y = 25000 \\ 300x + 400y = 27000 \end{cases}$$ Multiply first by 4 and second by 3: $$1600x + 1200y = 100000$$ $$900x + 1200y = 81000$$ Subtract second from first: $$700x = 19000 \implies x = \frac{19000}{700} = 27.14$$ Plug back into first: $$400(27.14) + 300y = 25000 \implies 10856 + 300y = 25000 \implies 300y = 14144 \implies y = 47.15$$ Check if low-grade constraint is met: $$200x + 500y = 200(27.14) + 500(47.15) = 5428 + 23575 = 29003 < 30000$$ Not enough low-grade oil, so use low-grade and medium-grade constraints: $$\begin{cases} 300x + 400y = 27000 \\ 200x + 500y = 30000 \end{cases}$$ Multiply first by 5 and second by 4: $$1500x + 2000y = 135000$$ $$800x + 2000y = 120000$$ Subtract second from first: $$700x = 15000 \implies x = \frac{15000}{700} = 21.43$$ Plug into first: $$300(21.43) + 400y = 27000 \implies 6429 + 400y = 27000 \implies 400y = 20571 \implies y = 51.43$$ Check high-grade constraint: $$400x + 300y = 400(21.43) + 300(51.43) = 8571 + 15429 = 24000 < 25000$$ Doesn't meet high-grade demand, so use high-grade and low-grade constraints: $$\begin{cases} 400x + 300y = 25000 \\ 200x + 500y = 30000 \end{cases}$$ Multiply second by 2: $$400x + 300y = 25000$$ $$400x + 1000y = 60000$$ Subtract first from second: $$700y = 35000 \implies y = 50$$ Plug back into first: $$400x + 300(50) = 25000 \implies 400x + 15000 = 25000 \implies 400x = 10000 \implies x = 25$$ Check medium-grade constraint: $$300x + 400y = 300(25) + 400(50) = 7500 + 20000 = 27500 \geq 27000$$ All constraints satisfied. ### Final answer: Run Refinery 1 for **25 days** and Refinery 2 for **50 days**. ### Minimum cost: $$C = 20000(25) + 25000(50) = 500000 + 1250000 = 1750000$$