A farmer wants to build a rectangular region, using a river as one side and some fencing as the other three sides. He has 1200 feet of fence which he can arrange to different dimensions. He creates the rectangular region with length $ L$ and width $ W$ to enclose the greatest area. Find $ L\plus{}W$.
1. **Set up the equation for the perimeter:**
   Since the farmer is using the river as one side of the rectangle, he only needs to fence the other three sides. Let $L$ be the length parallel to the river and $W$ be the width perpendicular to the river. The total length of the fence used is given by:
   \[
   2L + W = 1200
   \]
   Solving for $W$ in terms of $L$:
   \[
   W = 1200 - 2L
   \]

2. **Set up the equation for the area:**
   The area $A$ of the rectangle is given by:
   \[
   A = L \times W
   \]
   Substituting $W$ from the previous equation:
   \[
   A = L \times (1200 - 2L) = 1200L - 2L^2
   \]

3. **Maximize the area function:**
   To find the maximum area, we need to find the critical points of the function $A(L) = 1200L - 2L^2$. This is a quadratic function, and it opens downwards (since the coefficient of $L^2$ is negative), so the maximum value occurs at the vertex. The vertex of a quadratic function $ax^2 + bx + c$ occurs at:
   \[
   L = -\frac{b}{2a}
   \]
   Here, $a = -2$ and $b = 1200$, so:
   \[
   L = -\frac{1200}{2(-2)} = \frac{1200}{4} = 300
   \]

4. **Find the corresponding width:**
   Using $L = 300$ in the equation $W = 1200 - 2L$:
   \[
   W = 1200 - 2(300) = 1200 - 600 = 600
   \]

5. **Calculate $L + W$:**
   \[
   L + W = 300 + 600 = 900
   \]

The final answer is $\boxed{900}$.