Notice that the height of the box is equal to 4 in. Therefore, the volume of the box is

Where l is the length, w is the width, and h is the height of the box.
Therefore, since the volume of the box is 1232 in^2

Solve the quadratic equation as shown below,
![\begin{gathered} \Rightarrow x=\frac{8\pm\sqrt[]{64+1232}}{2}=(8\pm36)/(2)\to\text{ x has to be positive because it is a length} \\ \Rightarrow x=22 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4uwpk66hvnx5vsgvngzanayk8bjm74inpp.png)
Thus, the answer is that the original length is equal to 30in and the original width is 22in.