3

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 $(L,W)=(24,12)$

Mike McMahon
  • 145
  • 5
  • 1
    The two equations you wrote down are right. The logic is very clear, there is no reason to doubt. Something mechanical went bad later. – André Nicolas Sep 05 '13 at 01:04
  • so am i to infer that the length of the fence could be either 12 or 36ft (since it can't really be negative feet) and then solve for the perimeter knowing that 1w*(1/2L)=(576/2) ? Or do i blame the builders? – Mike McMahon Sep 05 '13 at 01:08
  • The equation solves nicely, positive answers, values of $W$ are $12$ and $32$. No need to worry about buying negative lengths of fence. – André Nicolas Sep 05 '13 at 01:24

1 Answers1

3

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:

  • $L = 9, W = 32$
  • $L = 24, W = 12$
Amzoti
  • 56,093