Question

The length of a rectangle is 6 inches longer than its width. What are the possible widths if the area of the rectangle is at least 667 square inches?

Answer 1

D,
`w\geq 23`

Explanation:

The length is 6 inches longer than it's width, therefore:

`l=w+6`

The area of the rectangle is AT LEAST 667 square inches, so:

`lw\geq 667`

Now, we have a system of two equations:

`\left \{ {{l=w+6} \atop {lw\geq 667}} \right.`

We want to isolate the width, so substitute the first equation into the second to eliminate the length.

`(w+6)w\geq 667`

Simplify:

`w^2+6w\geq 667`

Since we now have a polynomial equation that needs to be factored, set the value of one side to 0.

`w^2+6w-667\geq 0`

Now factor the equation:

`(w-23)(w+29)\geq 0`

Taking each factor individually, we find the following:

`w\geq 23\\w\geq -29`

Since a negative width doesn't make any sense, we can eliminate that solution, and we are left with the correct answer!

`w\geq 23`