Answer
See below
Work Step by Step
The maximum area = $300$.
For determining the area of the frame, we have: $(2x+10)^2-10^2$
Thus, we obtain: $(2x+10)^2-10^2=300\\
\Leftrightarrow 4x^2+40x+100-100=300\\\Leftrightarrow4x^2+40x-300=0\\ \Leftrightarrow (2x+30)(2x-10)=0\\ \Leftrightarrow 2x+30=0 \lor 2x-10=0\\
\Leftrightarrow x=-15 \lor x=5$
Since the width of the frame cannot be negative, the solution for this problem is $x=5$.
Hence, the width is 5 inches.
