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 832 square inches?

Answer 1

let W be the width and L be the length, then L=W+6, and so area is\[w(w+6) \ge 832\]or\[x^2+6w-832 \ge0\]which factors to\[(w-26)(w+32) \ge0\]yielding roots W=-32 union W=26.

Now looking at a rough sketch of the parabola
we see that it is positive for values \[w \le-32 \cup w \ge 26\]of course negative values of W do not make sense so the possible widths are those that are greater than or equal to 26 up to ____.

(there must be a max; let me think about that)