A lifeguard needs to rope off a rectangular swimming area in front of Long Lake Beach, using 1900 yd of rope and floats. What dimensions of the rectangle will maximize the area? What is the maximum area? (Note that the shoreline is one side of the rectangle.) Let $x$ be the length of a side of the rectangle perpendicular to the shoreline. Write the objective function for the area in terms of $x$. \[ \mathrm{A}(\mathrm{x})= \] (Type an expression using $x$ as the variable.)

Step 1 ：Let $x$ be the length of a side of the rectangle perpendicular to the shoreline. The total length of the rope will be used for the other three sides. This means that the perimeter of the rectangle is $2x + y = 1900$, where $y$ is the length of the side parallel to the shoreline.

Step 2 ：We can express $y$ in terms of $x$ as $y = 1900 - 2x$.

Step 3 ：The area of a rectangle is given by the product of its length and width, so the area $A$ in terms of $x$ is $A(x) = x * (1900 - 2x)$.

Step 4 ：Taking the derivative of the area function, we get $A'(x) = 1900 - 4x$.

Step 5 ：Setting the derivative equal to zero, we find the critical point $x = 475$. This is the value of $x$ that maximizes the area of the rectangle.

Step 6 ：Substituting this value back into the area function, we find the maximum area $A = 475 * (1900 - 2*475) = 451250$.

Step 7 ：Final Answer: The dimensions of the rectangle that will maximize the area are $475$ yd perpendicular to the shoreline and $950$ yd parallel to the shoreline. The maximum area is \(\boxed{451250}\) square yards.