A square always has the largest area given any single possible perimeter. For example with 20 ft of fencing for 4 sides, it is best to put 5 ft per side for an area of 25 ft^2. Nothing else gives you as great an area. Unfortunately this problem has a twist: there's no 4th side requirement for the perimeter. That changes things up a bit. Honestly I would resort to calculus here because it's just easy that way.
The perimeter is 2w + l = 120. The area is lw. Thus in terms of width, you have Area = l*w = (120-2w)*w = 120w - 2w^2 = 2w(60 - w). That's your area in terms of the width. You need to figure out when that number is the largest. Notice that the second term is squared and the first is linear (first power of w). That means that when w gets larger, the area gets SMALL. By searching for the zeros of this equation, you can get a good idea of where is is large and where it is small.
When you factor out the 2w, you see that the zeros are at w = 0 and w = 60. That means you have NO area when the width is 0 (duh) and when the width is 60 (because if each width is 60 then you've used up your 120 meters and you're left with 0 for the length). Somewhere in the middle of that is where it peaks - you can see that before w = 0 it is negative, between 0 and 60 it is positive, and after 60 it is negative.
Of course you can only have values between 0 and 60 or else it just wouldn't work in this world. If you're clever you'll notice that since this is a parabolic function, the origin of the parabola is exactly between the two zeros. That means at w = 30, you'll have your maximum.
With a width of 30, you have a length of 60, and this corresponds to an area of 1800 m^2. The answer is 30.
Using actual calculus you have A = 120w - 2w^2. The derivative is 120 - 4w. Solving for the zeros, you get w = 30. That's your maximum. Easy, huh?
