1. **Problem statement:** We want to maximize the area $A(x)$ of a rectangular plot with one side along a river, where only three sides are fenced with 800 meters of fencing wire. 2. **Define variables and formula:** Let $x$ be the length of the two sides perpendicular to the river, and the side along the river is $800 - 2x$ (since fencing is used on the other three sides). The area function is: $$A(x) = x(800 - 2x)$$ 3. **Find the derivative:** To find the maximum area, differentiate $A(x)$ with respect to $x$: $$A'(x) = 800 - 4x$$ 4. **Set derivative to zero to find critical points:** $$800 - 4x = 0 \implies 4x = 800 \implies x = 200$$ 5. **Check endpoints and critical point:** - At $x=0$, $A(0) = 0$ - At $x=400$, $A(400) = 400(800 - 800) = 0$ - At $x=200$, $A(200) = 200(800 - 400) = 200 \times 400 = 80,000$ 6. **Conclusion:** The maximum area is $80,000$ square meters when $x=200$ meters. The dimensions are: - Two sides perpendicular to the river: $200$ m each - Side along the river: $800 - 2(200) = 400$ m Thus, the best fencing plan encloses a rectangular plot of dimensions $200$ m by $400$ m with maximum area $80,000$ m².