A rectangular photo frame has outer length \(24\) and breadth \(18\). After removing a uniform border, the inner area is \(280\). What is the border width?

Correct answer: \(2\,\text{m}\)

Let the border width be \(w\). Then inner dimensions are \(24-2w\) and \(18-2w\). So
\[ (24-2w)(18-2w)=280.\]
Expand:
\[ 432-48w-36w+4w^2=280\]
\[ 4w^2-84w+152=0.\]
Divide by \(4\): \(w^2-21w+38=0\). Solving:
\[ w=\frac{21\pm\sqrt{21^2-4\cdot 38}}{2}=\frac{21\pm\sqrt{441-152}}{2}=\frac{21\pm\sqrt{289}}{2}=\frac{21\pm 17}{2}.\]
So \(w=19\) or \(w=2\). If \(w=19\), then \(24-2w<0\), not possible for an actual frame. Hence \(w=2\,\text{m}\).