Chapter 7.6, Problem 36E

To determine

To calculate: The length and width of the invitation letter and to determine its compatibility with the envelope.

Answer to Problem 36E

Length of invitation $=5inches$

Width of invitation $=3icnhes$

Length of envelope= $5\frac{1}{8}inches$

Width of envelope $=3\frac{5}{8}inches$

Invitation will fit inside envelope without being folded.

Explanation of Solution

Given information: Length of the letter is $1inch$ is less than twice its width, its area is 15 $sq.inches$ . Length of envelope is $5\frac{1}{8}inches$ and width of envelope is $3\frac{5}{8}inches$ .

Calculation:

Area of rectangle $=length\times width-(i)$

Given area $=15sq.m.$

Letting $length=l$ and $width=w$ ,

  $l=2(w)-1-(ii)$ (Given)

Putting $(ii)$ in $(i)$

  $\begin{array}{l}(2w-1)(w)=15\\ \Rightarrow 2w^2-w-15=0\end{array}$

Writing the equation with left side factored and solving for $\text{'}w\text{'}$ ,

  $\begin{array}{l}\Rightarrow 2w^2-6w+5w-15=0\\ \Rightarrow 2w(w-3)+5(w-3)=0\\ \Rightarrow (2w+5)(w-3)=0\\ \Rightarrow w=-\frac{5}{2},3\end{array}$

As width can’t be negative, $\therefore width=3inches$

Putting above value of width in equation $length\times width=15$ ,

  $\begin{array}{l}l\times 3=15\\ \Rightarrow l=5\end{array}$

  $\Rightarrow length=5inches$

Length of envelope= $5\frac{1}{8}inches$ (Given)

Clearly, length and width of envelope is more than length of invitation, $\therefore$ invitation will fit in envelope without being folded.