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Topic: Can someone show me the work for this?
Replies: 22   Last Post: Jun 6, 2005 4:01 AM

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Posts: 27
Registered: 1/29/05
Re: Can someone show me the work for this?
Posted: Jun 4, 2005 9:02 PM
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On Sun, 05 Jun 2005 00:10:21 GMT, "Bill" <xxx@yy.zz> wrote:

>"Darrell" <> wrote in message

>> "Peter Webb" <> wrote in message
>> news:42a197b3$0$5744$

>>> "tlew1234" <> wrote in message

>>>> Hi All,
>>>> I hope someone can help me with this.
>>>> Suppose you wanted to construct a fence around a garden plot in the form
>>>> of a rectangle. On the neighbor?s side it?s going to need heavy-duty
>>>> fencing that costs $2.00 per foot. The other three sides can be made of
>>>> standard fencing material that costs $1.20 foot. You have $200 to spend.
>>>> Question 1 How many different rectangles would it be possible to enclose
>>>> for $200?

>>> The first part of this is a setup for an extremely easy Calculus question.
>>> The actual question however is stupid. As somebody else said, there are an
>>> uncountable number of different rectangles.

>> It is certainly not stupid. It is good to know, from a mathematical
>> perspective, the answer to such a question. Why do you think it's a stupid
>> question?

>I'm sure someone will correct me if I'm wrong but to me it depends on what
>world you live in. If you live in a world of logic and take the words
>literally then there are an uncountable number of solutions.
>But if you live in the world of common sense there seems to be only one
>solution. (And everybody forgets the possibility of zero solutions.) For there
>to be more than one solution the property line on the neighbor's side would
>have to be L shaped or something more complicated than that. But you would
>have to cover that whole side with expensive fenceing (it said side, not
>neighbor's portion) - but the portion that is out side the neighbors property
>line seems to be no different then the others. So why would you need to cover
>the whole side with the more expensive fence?


You seem to be reading an extra constraint into the original question.
You seem to think that the side of the garden on the boundary with the
neighbor is "fixed" -- presumably at the length of that boundary. I
don't see that as given. The question is about fencing the garden, not
the neighbor. I think the Q allows that the garden size along the
boundary is any length (up to the length of the boundary, at least).
It seems to me to be "normal" that the size of the garden might be
some portion of the total yard (boundary).


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