Max Area Problem: Find Rectangle w/ Largest Area (400 yrs Fencing)

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David aims to enclose a rectangular area using 400 yards of fencing. The area A can be expressed as a function of width w, leading to the equation A = w(200 - w) after substituting the length in terms of width. The maximum area occurs when w is 100 yards, resulting in a maximum area of 10,000 square yards. While some participants suggest using derivatives for optimization, others recommend completing the square as an alternative method. The discussion includes light-hearted banter about the implications of using "years" in the context of fencing.
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1. David has 400 yrs of fencing and wishes to enclose a rectangular area. a) express the area A of the rectangle as a function of the width w of the rectangle. b) For what value of w is the area largest. c) What is the maximum area?



2. A=xy... My teacher does not want us to use derivatives.



3. I can't even being to attempt it.
 
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You got A=Lw and you want the area in terms of w, so the next step is to figure out what L is (in terms of w) and sub it into A=Lw.

Hint: use the perimeter formula
 
So does that mean L=400-w?
 
Cstreet09 said:
So does that mean L=400-w?

No. Draw a picture and label all 4 sides L or w. What is it that equals 400?
 
It may help to think of it this way:

What is greater?

(x-n)(x+n)
or
x2

To help, what is another way of writing the first one?


What's to say you can't use optimisation (derivatives) to check your answer?
 
There would be no reason to use derivatives. The function to be optimized is a quadratic and completing the square works nicely.
 
Cstreet09 said:
David has 400 yrs of fencing
That's going to take him a very long time!:wink:
 
:wink: He probably means light-years! :biggrin:
 
  • #10
Nah, I would have thought dark years
 
  • #11
ah … dark time

that mysterious phenomenon which is generally interspersed among ordinary time, but occasionally is more concentrated, causing effective time to pass more slowly and more heavily. :wink:
 
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