1. **Problem statement:** A rectangular pen is to be built with 1200 m of fencing, divided into three parts by two parallel partitions. We need to find the maximum possible area of the pen. 2. **Define variables:** Let $x$ be the length of the pen and $y$ be the width. 3. **Fencing constraint:** The total fencing includes the two lengths, four widths (two outer widths plus two partitions), so: $$2x + 4y = 1200$$ 4. **Express $x$ in terms of $y$:** $$2x = 1200 - 4y \implies x = 600 - 2y$$ 5. **Area function:** The area $A$ is: $$A = x \times y = (600 - 2y) y = 600y - 2y^2$$ 6. **Maximize area:** Take derivative with respect to $y$: $$\frac{dA}{dy} = 600 - 4y$$ Set derivative to zero for critical points: $$600 - 4y = 0 \implies y = 150$$ 7. **Find $x$ at $y=150$:** $$x = 600 - 2(150) = 600 - 300 = 300$$ 8. **Maximum area:** $$A_{max} = 300 \times 150 = 45000$$ --- **Part b:** If each side must be at least 200 m long: 9. **Check constraints:** $$x \geq 200, \quad y \geq 200$$ 10. **Check if $y=150$ satisfies $y \geq 200$:** No, so $y$ must be at least 200. 11. **Calculate $x$ when $y=200$:** $$x = 600 - 2(200) = 600 - 400 = 200$$ 12. **Area at $x=200$, $y=200$:** $$A = 200 \times 200 = 40000$$ 13. **Conclusion:** The maximum area decreases to 40000 when each side must be at least 200 m long.