Question 892960
w for width
L for length
A for area
A=4500
L=8+w
Using your choice of 1*L to serve as one side of the rectangle region to be fenced, you choose L+2w for the length of fencing to use, c.
c for length of fencing to use.


Perimeter of the rectangle region is 2w+2L, but your length of fencing is c=L+2w.

{{{A=wL}}}
{{{A=w(w+8)}}}
{{{w^2+8w=A}}}
{{{w^2+8w-A=0}}}------ One unknown variable, w, and one known variable, A as a given constant.
You have the choice to substitute for A now and if factorable quadratic, then factor the quadratic expression to solve for w; or keep going in symbolic form and substitute for A later.  Seeing so many different factorizations for 4500, my work will be done first in symbolic form.


{{{w=(-8+- sqrt(64-4A))/2}}}
{{{w=(-8+2sqrt(16-A))/2}}}-------   The PLUS form will be what is needed
{{{highlight_green(w=-4+sqrt(16-A))}}}
{{{w=-4+sqrt(16+4500)}}}
{{{highlight(w=-4+sqrt(4516))}}}, keeping like this for now.


{{{wL=4500}}}
{{{L=4500/w}}}
{{{L=4500/(-4+sqrt(4516))}}}
If you rationalize the denominator and simplify (not showing those steps here),
{{{highlight(L=(4500+1125sqrt(4516))/1133)}}}----not simplifiable further.


The amount of fence material c=L+2w
{{{highlight(c=(4500+1125sqrt(4516))/1133)+2sqrt(4516)-8)}}}
You can then compute this as a decimal approximation if needed.
You could also first compute L and w and then find the value for c.