1. **Problem statement:** We have a rectangular enclosure divided by a fence parallel to two of its sides, using a total of 600 meters of fencing. We want to find the dimensions that maximize the enclosed area. 2. **Define variables:** Let the length of the rectangle be $L$ and the width be $W$. Since the fence divides the rectangle parallel to two sides, there are 3 lengths and 2 widths of fencing used. 3. **Fence length equation:** Total fence used is 600 meters, so $$3L + 2W = 600$$ 4. **Area formula:** The area $A$ of the rectangle is $$A = L \times W$$ 5. **Express $W$ in terms of $L$:** From the fence equation, $$2W = 600 - 3L \implies W = \frac{600 - 3L}{2}$$ 6. **Area as a function of $L$:** Substitute $W$ into the area formula, $$A(L) = L \times \frac{600 - 3L}{2} = 300L - \frac{3}{2}L^2$$ 7. **Maximize area:** Take the derivative of $A(L)$ with respect to $L$ and set it to zero, $$\frac{dA}{dL} = 300 - 3L = 0 \implies 3L = 300 \implies L = 100$$ 8. **Find $W$:** Substitute $L=100$ back into the expression for $W$, $$W = \frac{600 - 3(100)}{2} = \frac{600 - 300}{2} = \frac{300}{2} = 150$$ 9. **Check second derivative:** $$\frac{d^2A}{dL^2} = -3 < 0$$ which confirms a maximum. 10. **Answer:** The dimensions that maximize the area are length $L = 100$ meters and width $W = 150$ meters.