Find the smallest perimeter and the dimensions for a rectangle with an area of $144 \mathrm{in}^{2}$ The smallest perimeter for a rectangle with an area of 144 in. ${ }^{2}$ is $48 \mathrm{in}$. (Simplify your answer.) The dimensions of this rectangle are in. (Simplify your answers. Use a comma to separate answers.)

Step 1 ：The area of a rectangle is given by the formula \(Area = length \times width\). Given that the area is 144 square inches, we need to find the length and width such that their product is 144 and their sum (which would be half the perimeter) is as small as possible.

Step 2 ：The length and width that minimize the perimeter would be when the rectangle is a square, because for a given area, a square has the smallest perimeter.

Step 3 ：So, we need to find the square root of 144 to get the length and width of the square.

Step 4 ：The perimeter of a rectangle is given by the formula \(Perimeter = 2 \times (length + width)\). Once we have the length and width, we can calculate the perimeter.

Step 5 ：Final Answer: The dimensions of the rectangle are \(12 \mathrm{in}\) by \(12 \mathrm{in}\) and the smallest perimeter is \(\boxed{48 \mathrm{in}}\).