Find the length and width of a rectangle whose perimeter is 42 feet and whose area is 108 square feet. The length (the longer side) is $\square$ feet and the width (the shorter side) is $\square$ square feet. (Simplify your answers.)

Step 1 ：We are given a rectangle with a perimeter of 42 feet and an area of 108 square feet. We need to find the length and width of the rectangle.

Step 2 ：We know that the perimeter of a rectangle is given by \(2 \times (Length + Width)\) and the area is given by \(Length \times Width\).

Step 3 ：Setting up the equations, we get \(2 \times (L + W) = 42\) and \(L \times W = 108\).

Step 4 ：Solving these equations, we get two possible pairs of values for L and W: (9, 12) and (12, 9).

Step 5 ：However, the length should be the longer side, so we choose (12, 9) as our solution.

Step 6 ：Final Answer: The length of the rectangle is \(\boxed{12}\) feet and the width is \(\boxed{9}\) feet.