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 \( 144 \mathrm{in}^{2}{ }^{2} \) is \( \square \mathrm{in} \).

To find the smallest perimeter of a rectangle with a given area, we can use the relationship between the area \( A \), the length \( l \), and the width \( w \) of the rectangle: \[ A = l \times w \] Given that the area \( A = 144 \, \text{in}^2 \), we have: \[ l \times w = 144 \] The perimeter \( P \) of a rectangle is given by: \[ P = 2l + 2w = 2(l + w) \] To minimize the perimeter, we can express \( w \) in terms of \( l \): \[ w = \frac{144}{l} \] Substituting this into the perimeter formula gives: \[ P = 2\left(l + \frac{144}{l}\right) \] To find the minimum perimeter, we can take the derivative of \( P \) with respect to \( l \) and set it to zero: \[ \frac{dP}{dl} = 2\left(1 - \frac{144}{l^2}\right) \] Setting the derivative equal to zero: \[ 1 - \frac{144}{l^2} = 0 \] Solving for \( l^2 \): \[ \frac{144}{l^2} = 1 \implies l^2 = 144 \implies l = 12 \] Now, substituting \( l = 12 \) back to find \( w \): \[ w = \frac{144}{12} = 12 \] Thus, both the length and width are \( 12 \, \text{in} \). Now we can calculate the perimeter: \[ P = 2(l + w) = 2(12 + 12) = 2 \times 24 = 48 \, \text{in} \] Therefore, the smallest perimeter for a rectangle with an area of \( 144 \, \text{in}^2 \) is \[ \boxed{48 \, \text{in}} \] The dimensions of the rectangle are \( 12 \, \text{in} \) by \( 12 \, \text{in} \).