 # 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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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
$$\displaystyle{r}={3}√{2}{A}$$