Well, we can denote L and W for the length and width respectively. Lets say the A is the area, we have: 1. A=(L × W) as well as 2. 2(L+W)=400. We rearrange the second equation to get 3. W=200-L. From this, we can see that 0<L<200. Substitute the third equation into the first to get A=(200L-L²). put this formula into the scientific calculator and you will find a parabola with a maximum. That would be the maximum area of the enclosed area. Alternatively, we can say that L is between 0 and 200 when the area equals 0. (The graph you find will be area against length). As the maximum is generally found halfway, we substitute 100 into the equation and we end up with 10000.
