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* w=22`

Then, substitute the expressions in the area:

`\begin{gathered} l* w=22 \\ (7+2w)* 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=(-7\pm√(7^2-4*\:2\left(-22\right)))/(2*\:2)`

Simplify:

`w=(-7\pm√(49+176))/(4)=(-7\pm√(225))/(4)=(-7\pm15)/(4)`

The solutions are:

`\begin{gathered} w=(-7+15)/(4)=(8)/(4)=2 \\ and \\ w=(-7-15)/(4)=-(22)/(4)=-(11)/(2) \end{gathered}`

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

Answer: 2 inches