Question

The length of a rectangle is 7 inches more than twice the width. If the area is to be at least 22 square inches,what are the possibilities for the width?The width is at least___ inches.

Answer 1

Given:

The length of a rectangle is 7 inches more than twice the width, this can be expressed by:

$l=7+2w$

Where:

l = length

w = width

And the area is 22 square inches, which is given by:

$A=l\times w=22$

Then, substitute the expressions in the area:

$\begin{gathered} l\times w=22 \\ (7+2w)\times w=22 \end{gathered}$

And solve for w:

$\begin{gathered} 7w+2w^2=22 \\ re\text{ order} \\ 2w^2+7w-22=22-22 \\ 2w^2+7w-22=0 \end{gathered}$

We obtain a quadratic equation, so we solve using the general formula for these equations:

a = 2

b = 7

c = -22

$w=\frac{-7\pm\sqrt{7^2-4\times\:2\left(-22\right)}}{2\times\:2}$

Simplify:

$w=\frac{-7\pm\sqrt{49+176}}{4}=\frac{-7\pm\sqrt{225}}{4}=\frac{-7\pm15}{4}$

The solutions are:

$\begin{gathered} w=\frac{-7+15}{4}=\frac{8}{4}=2 \\ and \\ w=\frac{-7-15}{4}=-\frac{22}{4}=-\frac{11}{2} \end{gathered}$

Since a length can't be negative, then the answer is w = 2.

Answer: 2 inches