[phoenirius]

Ms. Brown has 200m of fencing with which she intends to construct a rectangular enclosure along a river where no fencing is needed. She plans to divide the enclosure into two parts to separate her sheep from her coats, two perfect squars with right angles, river at upper portion of diagram.
What should be the dimensions of the enclosure if its area is to be a maximum?

Where do I start? How do I begin this equation?

[NeedKarma]

I always try to diagram stuff like this. I see the problem as looking like this:



Got to imagine that those are two perfect squares. If they are then all of those segments will be of equal length. I count 6 segments that must total 200m so...

6x = 200
x = 200/6
x = 33.33

That's just one segment. You can take it from there or another expert can correct me.

[CaptainForest]

NeedKarma , I count 7 segments (dont you need one on the river so they don't jump into the river)

so

7x = 200
x = 28.57

[phoenirius]

I count 7 segments too. So does that mean 7x=200 is my algebraic equation for this question? Isn't there more too it?

[NeedKarma]

I was going with this part of the word problem:

a rectangular enclosure along a river where no fencing is needed.

I assumed they said that because the river is the segment.

[phoenirius]

Sorry, your right. I wasn't paying attention.

[CaptainForest]

I was going with this part of the word problem:

Quote:
a rectangular enclosure along a river where no fencing is needed.

I assumed they said that because the river is the segment.

Ahh, yes. I think you are right.

[s_cianci]

Your question is unclear. Do you want to divide the enclosure into two square regions or do you want the maximum area possible, regardless of whether the enclosures are squares?

[s_cianci]

NeedKarma , I count 7 segments (dont you need one on the river so they dont jump into the river)

so

7x = 200
x = 28.57

But your problem states that no fencing is needed along the river.

[letmetellu]

I don't know the formula but I know how I would construct the fence to get the maximum area. Just for ease of explaining I would build a three sided square area bordering the river, and inside that area I would build another two sided square, using one side of the first square and also using the river for the forth side. I would build the small area in either corner of the large square that adjoins the river.

[CaptainForest]

But your problem states that no fencing is needed along the river.

I know. That is why I retracted and agreed that NeedKarma was right.

[phoenirius]

I stated that it was(the river) on the upper part of the two perfect squares, meaning it isn't on the sides which then elminates 2sides right? Leaving 5.
This still isn't helping though. Is my complete aglebraic expression just 5x=200 and carried through??

[phoenirius]

Your question is unclear. Do you want to divide the enclosure into two square regions or do you want the maximum area possible, regardless of whether or not the enclosures are squares?

The information I gave here is all the information I was gave. I'm assuming they want the maximum area possible while at the same time separating the two areas, so there will be two separate squares but the maximum amount of fencing needs to be found for each one.

[CaptainForest]

I stated that it was(the river) on the upper part of the two perfect squares, meaning it isn't on the sides which then elminates 2sides right? Leaving 5.
This still isn't helping though. Is my complete aglebraic expression just 5x=200 and carried through???

Yes, it would be 5x=200 since if they both hit the water it leaves only 5 segments

[NeedKarma]

Yes, it would be 5x=200 since if they both hit the water it leaves only 5 segments

Ah yes, I missed that part. It certainly makes for some easier math.:)

[reinsuranc]

So we have found two ways to construct these areas. The first way uses six equal lengths of fence, 6x=200, x=33.3. The second way uses five equal lengths of fence, 5x=200, x=40.

The first way creates two squares with total area (33.3 x 33.3) + (33.3 x 33.3) = 2218 rounded. The second way creates two squares with total area (40 x 40) + (40 x 40) = 3200.

Thus the second way gives the larger area.

Now, did we decide that it would be OK, to put one square inside another? If so, suppose we use 4 feet to form a small 1 by 1 square, and the remaining 196 feet to form three sides (with the river being the fourth side) of a square surrounding the smaller square. 196/3 = 65.3. So now the larger square is 65.3 x 65.3 = 4264 of area rounded. That's an even larger area.

We could make that smaller square infintely small, couldn't we? How about an inch by an inch square? That's not too practical, but that would leave nearly the entire 200 feet to form 3 sides, each of length 66 2/3, and the area would be 66 2/3 x 66 2/3 = 4445 of area rounded.

[phoenirius]

We can't put squares inside each other and we can't have 5 sides and 6 sides and 7 sides. I appreciate everyone's help but we're still off from what the question is asking, I've never been so confused in my life.

I completely understand why you would make it simplified and say 5x=200 and x=40m but I need a complete algebraic equation finalizing the question in the form y=a(x-p)^2+q

So how would I use the area formula and the perimeter formula to solve this into that final form?

Can somebody please just help me figure this out, give me the steps, something to work with. I'm being told that I'm not solving for 5unknown sides but 3 and that I need 200=l+3w broke down to l=3w-200... I need all the previous steps, at least how to get them. I really need help on this. I'd really really appreciate it.

[letmetellu]

Draw us a picture of what you want the squares to look lile.