Question 475831
Assuming that the developer fences three sides of a rectangle, we can define dimensions x and y such that


2x + y = 336


Area = xy


By the AM-GM inequality (Google this if you don't know what it is),


*[tex \LARGE \frac{2x + y}{2} \ge \sqrt{2xy}]


*[tex \LARGE \frac{336}{2} \ge \sqrt{2xy}]


*[tex \LARGE 168 \ge \sqrt{2xy}]


*[tex \LARGE 168^2 \ge 2xy \Rightarrow xy \le \frac{168^2}{2} \Rightarrow Area \le 14112]


The largest possible area is 14112 ft^2, which occurs in the equality case (2x = y).