1. **State the problem:** A rancher has resources: a feedlot for 200 steers, 100 hectares of land for sheep or barley. Each steer needs 0.5 tonnes barley. Profit per steer (excluding barley cost) is 100. Sheep yield 280 per hectare. Barley yields 4 tonnes/hectare, costs 80/hectare to produce, sells for 140/tonne, or can be fed to steers. Extra barley can be bought at 150/tonne. Goal: allocate land and barley to maximize profit. 2. **Define variables:** Let $x$ = hectares used for barley (0 \leq x \leq 100), $y$ = hectares used for sheep (0 \leq y \leq 100), $z$ = number of steers fattened (0 \leq z \leq 200). 3. **Constraints:** - Land: $x + y \leq 100$ - Barley needed for steers: $0.5z$ tonnes - Barley produced: $4x$ tonnes - Barley shortfall (if any) bought: $b = \max(0, 0.5z - 4x)$ tonnes 4. **Profit components:** - Profit from steers excluding barley cost: $100z$ - Cost of barley produced: $80x$ - Revenue from selling barley (if any left after feeding steers): $140(4x - 0.5z)$ if $4x > 0.5z$, else 0 - Cost of buying extra barley: $150b$ - Profit from sheep: $280y$ 5. **Total profit function:** $$ P = 100z + 280y - 80x + 140 \max(0, 4x - 0.5z) - 150 \max(0, 0.5z - 4x) $$ 6. **Analyze cases:** - Case 1: $4x \geq 0.5z$ (enough barley produced) $$P = 100z + 280y - 80x + 140(4x - 0.5z) = 100z + 280y - 80x + 560x - 70z = (100z - 70z) + 280y + (560x - 80x) = 30z + 280y + 480x$$ - Case 2: $4x < 0.5z$ (need to buy barley) $$P = 100z + 280y - 80x - 150(0.5z - 4x) = 100z + 280y - 80x - 75z + 600x = (100z - 75z) + 280y + (600x - 80x) = 25z + 280y + 520x$$ 7. **Maximize profit under constraints:** - Land: $x + y \leq 100$ - Steers: $z \leq 200$ - Barley and steers relation for cases 8. **Check Case 1 (enough barley produced):** Condition: $4x \geq 0.5z \Rightarrow z \leq 8x$ Maximize $P = 30z + 280y + 480x$ with $x + y \leq 100$, $z \leq 200$, $z \leq 8x$ - To maximize $P$, increase $y$ and $x$ as they have positive coefficients. - Max $y = 100 - x$ - Max $z = \min(200, 8x)$ Try $x=25$, then $y=75$, $z=200$ (since $8*25=200$): $$P = 30*200 + 280*75 + 480*25 = 6000 + 21000 + 12000 = 39000$$ Try $x=20$, $y=80$, $z=160$: $$P = 30*160 + 280*80 + 480*20 = 4800 + 22400 + 9600 = 36800$$ Try $x=30$, $y=70$, $z=200$ (check $z \leq 8x$? $8*30=240 > 200$ okay): $$P = 30*200 + 280*70 + 480*30 = 6000 + 19600 + 14400 = 40000$$ Try $x=33.33$, $y=66.67$, $z=200$: $$P = 30*200 + 280*66.67 + 480*33.33 = 6000 + 18667.6 + 15998.4 = 40666$$ Try $x=37.5$, $y=62.5$, $z=200$: $$P = 6000 + 17500 + 18000 = 41500$$ Try $x=40$, $y=60$, $z=200$: $$P = 6000 + 16800 + 19200 = 42000$$ Try $x=42.5$, $y=57.5$, $z=200$: $$P = 6000 + 16100 + 20400 = 42500$$ Try $x=45$, $y=55$, $z=200$: $$P = 6000 + 15400 + 21600 = 43000$$ Try $x=50$, $y=50$, $z=200$: $$P = 6000 + 14000 + 24000 = 44000$$ Try $x=55$, $y=45$, $z=200$: $$P = 6000 + 12600 + 26400 = 45000$$ Try $x=60$, $y=40$, $z=200$: $$P = 6000 + 11200 + 28800 = 46000$$ Try $x=65$, $y=35$, $z=200$: $$P = 6000 + 9800 + 31200 = 47000$$ Try $x=70$, $y=30$, $z=200$: $$P = 6000 + 8400 + 33600 = 48000$$ Try $x=75$, $y=25$, $z=200$: $$P = 6000 + 7000 + 36000 = 49000$$ Try $x=80$, $y=20$, $z=200$: $$P = 6000 + 5600 + 38400 = 50000$$ Try $x=83.33$, $y=16.67$, $z=200$: $$P = 6000 + 4667.6 + 39998.4 = 50666$$ Try $x=85$, $y=15$, $z=200$: $$P = 6000 + 4200 + 40800 = 51000$$ Try $x=90$, $y=10$, $z=200$: $$P = 6000 + 2800 + 43200 = 52000$$ Try $x=95$, $y=5$, $z=200$: $$P = 6000 + 1400 + 45600 = 53000$$ Try $x=100$, $y=0$, $z=200$: $$P = 6000 + 0 + 48000 = 54000$$ 9. **Check Case 2 (need to buy barley):** Condition: $4x < 0.5z \Rightarrow z > 8x$ Profit: $$P = 25z + 280y + 520x$$ Max $z=200$, max $y=100 - x$, $z > 8x$ means $x < 25$ Try $x=24$, $y=76$, $z=200$: $$P = 25*200 + 280*76 + 520*24 = 5000 + 21280 + 12480 = 38760$$ Try $x=20$, $y=80$, $z=200$: $$P = 5000 + 22400 + 10400 = 37800$$ Try $x=10$, $y=90$, $z=200$: $$P = 5000 + 25200 + 5200 = 35400$$ 10. **Conclusion:** Maximum profit is in Case 1 with $x=100$, $y=0$, $z=200$: - Use all 100 hectares for barley - Fatten all 200 steers - No sheep - Profit = 54000 **Final answer:** The farmer should use all 100 hectares to grow barley, fatten all 200 steers, and not run sheep to maximize profit of 54000.