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?

Question

asked Feb 10, 2023 231k views

1 Answer

Anshuma · Answer 1 · 2023-02-16T02:57:00+0000

Answer:

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!

$w\geq 23$