A wire of length r feet is bent into a rectangle whose width is 2 time

opatovaL 2021-08-14 Answered
A wire of length r feet is bent into a rectangle whose width is 2 times its height. Write the area A of the rectangle as a function of the wire's length r. Write the wire's length r as a function of the area A of the rectangle (note A not a).

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Expert Answer

Fatema Sutton
Answered 2021-08-15 Author has 6047 answers

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 r in terms of A:
\(\displaystyle{18}{A}={r}^{{2}}\)
\(\displaystyle√{18}{A}={r}\)
or
\(\displaystyle{r}={3}√{2}{A}\)

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