1. **State the problem:** A farmer wants to enclose two adjacent rectangular plots for sheep and one for cattle along a river, using 240 yards of fencing. The river forms one side, so no fencing is needed there. We want to find the largest total area that can be enclosed. 2. **Define variables:** Let $x$ be the length of one rectangular plot and $w$ be the width of one rectangular plot. Since there are two adjacent plots for sheep, their combined length is $2x$, and the width is $2w$. 3. **Fencing constraint:** The fencing is needed for the three widths and three lengths not along the river. The river forms one side, so fencing is needed for the opposite side and the internal fences. The total fencing used is: $$\text{Fencing} = 3w + 2x$$ Given 240 yards of fencing: $$3w + 2x = 240$$ 4. **Express $w$ in terms of $x$:** $$3w = 240 - 2x$$ $$w = \frac{240 - 2x}{3}$$ 5. **Area function:** The total area $A$ is the sum of the areas of the two sheep plots and one cattle plot. Since the two sheep plots are adjacent along length $x$, total length is $2x$ and width is $w$, and the cattle plot has length $x$ and width $w$. Total area: $$A = 2(xw) + xw = 3xw$$ Substitute $w$: $$A = 3x \times \frac{240 - 2x}{3} = x(240 - 2x) = 240x - 2x^2$$ 6. **Maximize area:** To find the maximum area, take the derivative of $A$ with respect to $x$ and set it to zero: $$\frac{dA}{dx} = 240 - 4x = 0$$ $$4x = 240$$ $$x = 60$$ 7. **Find $w$:** $$w = \frac{240 - 2(60)}{3} = \frac{240 - 120}{3} = \frac{120}{3} = 40$$ 8. **Calculate maximum area:** $$A = 3 \times 60 \times 40 = 7200$$ **Final answer:** The largest total area that can be enclosed is **7200 square yards**.