I have a rectangle divided like so
----------------------
| | |
| | |
| | | W
| | |
| | |
----------------------
L L
- The total area of both regions combined is 576
- the total perimeter (including the inner fence) is 132
From this information we need to determine the sub region's perimeter....and this is where I am stuck!
We are currently working on quadratic equations at the moment, so i need to get this into the form of $ax^2+bx+c=0$
My initial thought is i need to intersect $2(L*W)=576$ and $ 3W + 4L = 132 $
$$2(L*W)=576$$ $$L*W=\frac{576}{2}$$ $$L*W=288$$
Then I take $3W + 4L = 132$ and do the following
$$3W+4L=132$$ $$4L=132-3W$$ $$L=\frac{132-3W}4$$ or $$L=33-.75W$$
I do my substitution
$$W(33-.75W)-288=0$$
simplified (use that term loosely)
$$-.75W^2+33W-288=0$$ $$-100(-.75W^2+33W-288=0)$$ $$75W^2+3300W-28800=0$$
translating to
$$-(-3300)\pm \sqrt{(-300)^2-(4*75*28800)}\over 2*75$$
My results were $(L,W)=(9,32)$ and $(L,W)=(24,12)$
We have:
$$L = \dfrac{576}{W}$$
Substituting $L$ into $3W + 2L = 132$, yields:
$$3W^2 - 132 W + 2* 576 = 0$$
This (use the quadratic) yields two positive results for $W$:
$$W = 12~~ \text{or}~~ 32$$
Note that if we had gotten a positive and negative, we would discard the negative.
Now find the two values of $L$.
You should get $(W, L) = (12, 48)$ or $(W, L) = (32, 18)$
So, yes, this problem results in two valid and distinct results, which you should actually verify.
Update
We have:
$$2L = \dfrac{576}{W} \rightarrow L = \dfrac{288}{W}$$
Substituting $L$ into $3W + 4L = 132$, yields:
$$3W^2 - 132 W + 1152 = 0$$
From this we arrive at: