Question

The area of the conference table in Mr. Nathan’s office must be no more than 175 ft2. If the length of the table is 18 ft more than the width, x, which interval can be the possible widths?

Answer 1

width=x
length=x+18
area=x(x+18),assuming table to be rectangular.,then
\[x \left( x+18 \right)\le 175\]
\[x ^{2}+18x \le 175\]
adding both sides (18/2)^2 i.e.,81
\[\left( x+9 \right)^{2}=175+81\]
\[\left| x+9 \right|\le \sqrt{226}\]
\[-\sqrt{226}\le x+9\le \sqrt{226}\]
\[-9 -\sqrt{226}\le x \le -9+\sqrt{226},\]
But x >0
\[Hence 0 < x \le -9-\sqrt{226}\]