The largest possible area for a given amount of material enclosing a rectangle will always be a perfect square with each side being one quarter of the amount of material.
The proof is:
M=2x+2y, solve for y
y=(M-2x)/2
A=xy, and using y from above...
A=(Mx-2x^2)/2
dA/dx=(M-4x)/2
d2A/dx2=-2
Since acceleration is always negative, when velocity is zero, it will be at an absolute maximum for A(x)
dA/dx=0 when M-4x=0, 4x=M, x=M/4
So the maximum area occurs when x=M/4, and from earlier:
y=(M-2x)/2 then use x=M/4 in this and get:
y=(M-M/2)/2
y=(2M-M)/4=M/4 so x=y=M/4, a perfect square.
So for your particular example...
M=2x+2y and M=100
100=2x+2y
50=x+y
y=50-x
Now:
A=xy, using y found above...
A=50x-x^2
dA/dx=50-2x
d2A/dx2=-2 so an absolute maximum occurs when dA/dx=0
dA/dx=0 when 50-2x=0, 2x=50, x=25...
and from earlier, y=50-x, using x from above y=50-25=25
so y=x=25
So the maximum area with 100 feet of fencing is 25^2=625ft^2
