June 2001 Question: 
A neighborhood group decided to use a vacant lot to build a park for all neighborhood kids to use. They planned and decided to make a basketball court and playground equipment. On one side of the lot, a large fence was already built, so they decided to fence in the rest of the lot using fencing that had been donated. The neighborhood group was given 155 meters of fencing. What is the largest rectangular lot which can be fenced to provide a sports complex for the neighborhood children?

June 2001 Answer:  
A great way to solve this is to try many different values, searching for a maximum number. We know that A = length x width (l x w). Initially, choose values and fill them in. For example: if the length was 105 meters, the width would be 25 (2 sides of the 25 total 50 left from 155 meters of fencing). So the area would be:

A = l x w
A = 105 x 25
A = 2,625

If the length was 95 meters, the width would be 30 meters. 

A = l x w
A = 95 x 30
A = 2,850

Keep trying new values. It may help to organize them in a chart.
Length Width
(2 sides)		 Perimeter of
new fencing		 Total
Area
105 m 25 m 155 m 2625 sq m
95 m 30 m 155 m 2850 sq m
85 m 35 m 155 m 2975 sq m
75 m 40 m 155 m 3000 sq m
65 m 45 m 155 m 2925 sq m

Looking at the table, it seems that 3000 sq. m is the greatest area. Before deciding on these values, however, try others with a length between 65 m and 85 m to see if any are larger.

Length Width
(2 sides)		 Perimeter of
new fencing		 Total
Area
83 m 36 m 155 m 2988 sq m
81 m 37 m 155 m 2997 sq m
79 m 38 m 155 m 3002 sq m
78 m 38.5 m 155 m 3003 sq m
77 m 39 155 m 3003 sq m
76 m 39.5 m 155 m 3002 sq m

Looking at this new data, the data seems to form a curve, with the highest possible value at 3003 sq. m, then falling on either side of that.

So, the largest are for the sports complex is 3003 sq. m if they build a fence with a length of 78 m and widths of 38.5 m or a length of 77 m and widths of 39 m.