To solve this question we will use the following diagram:
The area of the big rectangle minus the area of the small rectangle is:
![14in*16in-48in^2=176in^2.](https://img.qammunity.org/2023/formulas/mathematics/college/4n1cvp9evydxova6uomojlmf0obz8kruo5.png)
Since the frame has a uniform width, then:
![16x+16x+(14-2x)x+(14-2x)x=176.](https://img.qammunity.org/2023/formulas/mathematics/college/3rv7bsrm9p2e6z1ky1fzsz7kfrjh8ujtwr.png)
Solving the above equation for x we get:
![\begin{gathered} 32x+28x-4x^2=176, \\ 15x-x^2-44=0. \\ x^2-15x+44=0, \\ (x-4)(x-11)=0. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5lzr3l9zkdkuaulmnm4r2toop9bu1p529p.png)
Then:
![x=4\text{ or x=11,}](https://img.qammunity.org/2023/formulas/mathematics/college/2szh88enrs8kxkacfe5ub06h2xdi0gavx0.png)
but 11 is not a solution with real meaning, therefore x=4.
Answer: The width is 4 in.