Answer:
A(max)  = 28322 yd²
Dimensions:
x  =  238 yd
y  = 119 yd
Step-by-step explanation:
Let  call "x" and "y" horizontal and vertical sides of the rectangle,  then we have:
The total area, sum of areas of the three small rectangles is:
A(r) = x*y
And the length to be fenced is
P = 2*x  +  2*y  * 2*y
P = 2*x  + 4*y     and    952  = 2*x + 4*y   ⇒  y  = ( 952 - 2*x) / 4
Total area as function of x s:
A(r)  =  x * y   ⇒  A(x) =  x* ( 952 - 2*x) / 4
A(x)  =  238*x - (1/2)*x²
Taking derivatives on both sides of the equation we get:
A´(x)  = 238  - x   ⇒  A´(x) = 0    ⇒  238  - x = 0
x = 238 yd
Therefore  y =  ( 952 - 2*x) / 4
y  = ( 952  - 2* 238 ) / 4
y  =  119 yd
Largest area is:
A(max) Â = Â y * x Â
A(max) Â = 238 * 119
A(max)  = 28322 yd²