Q. 89 - Precalculus Enhanced with Graphing Utilities

Question

Geometry: Find formulas for the length l and width w of a rectangle in terms of its area A and perimeter P.

Step-by-Step Solution

Answer

The formula for length and width is $l=\frac{P+\sqrt{P^2-16A}}{4}$ and $w=\frac{P-\sqrt{P^2-16A}}{4}$.

Step 1. Given information

Area of the rectangle is given by $A=lw$.

Perimeter of the rectangle is $P=2(l+w)$.

So the system of equations is :

$$\begin{gathered} A=lw \\ P=2(l+w) \end{gathered}$$

Step 2. Solve the system of equations.

From first equation,

$$\begin{gathered} \frac{A}{w}=\frac{lw}{w} \\ \frac{A}{w}=l \\ l=\frac{A}{w} \end{gathered}$$

Substitute $l=\frac{A}{w}$ in $P=2(l+w)$.

$$\begin{gathered} P=2(\frac{A}{w}+w) \\ =2\left(\frac{A+w^2}{w}\right) \end{gathered}$$

Step 3. Solve the equation P = 2 ( A + w 2 w ) for w .

We get

$$\begin{gathered} Pw=2(A+w^2) \\ Pw=2A+2w^2 \\ 0=2w^2-Pw+2A \\ 2w^2-Pw+2A=0 \end{gathered}$$

Using the quadratic formula,

$$\begin{gathered} w=\frac{P\pm \sqrt{{(-P)}^2-4\left(2\right)\left(2A\right)}}{2\left(2\right)} \\ =\frac{P\pm \sqrt{P^2-16A}}{4} \end{gathered}$$

Step 4. Find an equation for l .

From first equation,

$$\begin{gathered} A=lw \\ w=\frac{A}{l} \end{gathered}$$

Substitute this in second equation.

$$\begin{gathered} P=2(l+\frac{A}{l}) \\ P=2\left(\frac{l^2+A}{l}\right) \\ Pl=2l^2+2A \\ 2l^2-Pl+2A=0 \end{gathered}$$

Using quadratic formula,

$l=\frac{P\pm \sqrt{P^2-16A}}{4}$

Step 5. Find the equation for l and w .

As the equation for both l and w is same, so assume

$l=\frac{P+\sqrt{P^2-16A}}{4}$ and

$w=\frac{P-\sqrt{P^2-16A}}{4}$.