The diagram of the problem is:
S is the length of the shorter side of the fence. L is the length of the longest side of the field.
We know that the perimeter of the rectangle is 800ft. This means:
[tex]2S+2L=800[/tex]
And the area:
[tex]A=SL[/tex]
The smaller rectangles will have dimensions:
The area is:
[tex]a=\frac{SL}{3}[/tex]
As we can see, if we maximize the area of the bigger rectangle "A", we are also maximizing the area of the smaller rectangles "a".
Then, we have two equations:
[tex]\begin{gathered} 2S+2L=800 \\ A=SL \end{gathered}[/tex]
We can solve for L in the first equation:
[tex]\begin{gathered} 2S+2L=800 \\ 2L=800-2S \\ L=400-S \end{gathered}[/tex]
Then substitute in the second:
[tex]A=S(400-S)[/tex]
Simplify:
[tex]A=400S-S^2[/tex]
This is a function of the area depending on the length of the shorter side of the rectangle:
[tex]A(S)=400S-S^2[/tex]
We can find the maximum of this function if we find the value where the derivative of this function is 0.
Let's differentiate:
[tex]A^{\prime}(S)=400-2S[/tex]
And now we find where A'(S) = 0:
[tex]\begin{gathered} 0=400-2S \\ 2S=400 \\ S=200 \end{gathered}[/tex]
We have found that the shorter side must have a length of 200ft to maximize the area. Let's find the length of the larger side:
[tex]L=400-200=200[/tex]
As expected, the quadrilateral which maximizes the area is the square. Thus, the dimensions of the field are 200ft x 200ft
