A farmer wishes to make a rectangular pen using an existing section of straight fence and 36 m of relocatable fencing materials. What is the largest possible area of the pen?
Is the existing section however long you need it to be?
Any rectangle must have a length L and a width W. For the perimeter to be $\displaystyle 32+x$, $\displaystyle 2L + 2W$ must be equal to $\displaystyle 32+x$.
We know that by the definition of a rectangle, the section of pre-existing fence will be equal in length to either the length or the width. So, we set $\displaystyle x $equal to $\displaystyle L$.
This gives us $\displaystyle 2L + 2W = 36+L$, or $\displaystyle L + 2W = 36$.
Also, we know that the maximum area of any rectangle is actually a rectangle with four equal sides. This is also known as a square.
With that knowledge, when looking for the maximum area we can say that $\displaystyle L = W$.
If $\displaystyle L = W$, and $\displaystyle L + 2W = 36$, then $\displaystyle W + 2W = 36$.
After that, it's just basic algebra.
$\displaystyle 3W = 36$
$\displaystyle W = 36/3$
We know the width, and since all the sides are equal, we can square it to find the area.
$\displaystyle (36/3)^2 = 144 m^2$
Perimeter must equal 36m: $\displaystyle 36=2(w+l)$
We want the largest area: $\displaystyle A=wl$
Solve the first equal for either length or width and plug it into the second equation. Calculus will give you the value for the largest area, but without knowledge of that math, you will need to use some intuition/trial and error...
$\displaystyle 18=w+l$
$\displaystyle A=(18-w)w$
$\displaystyle A=18w-w^2$
$\displaystyle w=9; l=9$
$\displaystyle A=81$