anonymous
  • anonymous
A rectangular field is enclosed by afence and divided into two smaller plots by a fence parallel to one of the sides. find the dimensions of the largest such field 1200m of fencing is avable
Calculus1
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@sasogeek @whpalmer4
anonymous
  • anonymous
@ganeshie8

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zepdrix
  • zepdrix
|dw:1433831834735:dw|Here is an illustration of what is being described.
zepdrix
  • zepdrix
We are given a constraint on the `perimeter`, amount of fencing. Err I guess perimeter is usually regarded as the distance around the outside. So we're talking about the perimeter plus this inner separator as well :p whatev \[\Large\rm P=3\ell + 2w\]That gives us our perimeter, ya? The constraint they gave us is that we have only 1200m of fencing to work with.\[\Large\rm 1200=3\ell + 2w\]
zepdrix
  • zepdrix
Use your constraint to write \(\Large\rm w\) in terms of \(\Large\rm \ell\), (or the other way around if you like). Then realize what you're trying to maximize, area. Area of this rectangle is going to be simply:\[\Large\rm A=\ell\cdot w\]
zepdrix
  • zepdrix
Using your constraint, replace your \(\Large\rm w\) in the area formula with some \(\Large\rm \ell\) stuff. From that point, since you now have your area function in terms of `one variable`, you can differentiate and look for critical points.
zepdrix
  • zepdrix
Hmm you've gone offline, hopefully that will help you get started at least :)
anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
sup :U

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