Question 760995
x for length of the whole area, and y for the width of the whole area.  Choose y longer than x, and so take the divider parallel to x.  


Accounting for fencing material, {{{3x+2y=200}}}.
Accounting for area, {{{A=xy}}}


Perimeter equation gives {{{2y=200-3x}}}
{{{y=100-(3/2)x}}}
Substitute into area equation.
{{{A=x(100-(3/2)x)}}}
{{{A=100x-(3/2)x^2}}}


A looks like a quadratic equation as a function, and indicates a parabola with a maximum ( openings downward).  Finding the zeros is very easy, only requiring back one step in deriving A(x).  The maximum A for AREA will occur directly in the middle of the two zeros of A(x).


ZEROS of A:
{{{0=x(100-(3/2)x)}}}
Either x=0
OR
100-(3/2)x=0
{{{100=3x/2}}}
{{{100*2/3=x}}}
{{{x=200/3}}}
Maximum Area occurs at {{{(0+200/3)/2=highlight(100/3=x)}}}

To find y,
{{{y=100-(3/2)x}}}
{{{y=100-(3/2)(100/3)=100-3*100/(2*3)=100-100/2}}}
{{{highlight(y=50)}}}



----------------------------------More about the area-----------
This area maximum is:
A(100/3)=100(100/3)-(3/2)(100/3)^2
A=10000/3-(3/2)*10000/9
A=1000/3-3*10000/18
A=6*10000/18-3*10000/18
A=3*10000/(3*6)=10000/6
A=1667 square feet