Question

Drew creates a table of ordered pairs representing the width and area of a dog pen.

A 2-column table with 5 rows. The first column is labeled width (feet) with entries 7, 8, 9, 10, 11. The second column is labeled area (square feet) with entries 77, 80, 81, 80, 77.
Which situation could describe Drew’s plans for the dog pen?

Drew has 36 feet of fencing and wants to represent the possible areas of the dog pen for possible widths.
Drew has a maximum of 81 feet of fencing to use around the pen, and the length must be twice the width.
Drew is building a rectangular dog pen with a length that is up to 4 feet more than the width and has a perimeter of at least 77 feet.
Drew wants to build a dog pen that is between 7 feet and 11 feet in width and uses, at most, 18 feet of fencing.

Answer 1

The first situation.

Reason: For each width w the length is A/w, giving lengths 11, 10, 9, 8, 7 for widths 7, 8, 9, 10, 11. So w + l = 18 and the perimeter 2(w + l) = 36 feet of fencing. The areas match A = w(18 − w), so Drew is modeling a rectangle with 36 ft of fencing.