The length of the wire is the perimeter of the rectangle with width ww and height hh:

\(2w+2h=r\)

Given that the width is 2 times its height, \(w=2h\), we write: \(2(2h)+2h=r\)

\(4h+2h=r\)

\(6h=r\)

\(\displaystyle{h}=\frac{{r}}{{6}}\)

which follows that:

\(\displaystyle{w}={2}⋅\frac{{r}}{{6}}=\frac{{r}}{{3}}\)

The area of the rectangle is:

\(A=wh\)

In terms of r,

\(\displaystyle{A}={r}{3}⋅{r}{6}\)

\(\displaystyle{A}=\frac{{r}^{{2}}}{{18}}\)

Solve for rr in terms of AA:

\(\displaystyle{18}{A}={r}^{{2}}\)

\(\displaystyle\sqrt{18A}={r}\)

or

\(\displaystyle{r}={3}\sqrt{2A}\)