anonymous
  • anonymous
A Rancher wants to enclose 2 rectangular area near a river, one for sheep and one for cattle. There is 240 yd. of fencing available. What is the largest total area that can be enclosed.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
This seems to be an optimization problem. Does the river count as one side that doesn't need to be fenced?
anonymous
  • anonymous
Yes it is an optimization problem. No I don't believe the river counts as a side that doesn't need to be fenced. It just says near a river so..I don't think so.
anonymous
  • anonymous
So, 2WidthCattle + 2LengthSheep + 2WidthCattle + 2 LengthSheep = 240 yards of fencing.

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anonymous
  • anonymous
And we know that length * width = area.
anonymous
  • anonymous
Ok, Im following so far
anonymous
  • anonymous
We also know that squares work the best, in terms of area.
anonymous
  • anonymous
So let's simplify our perimeter equation. If all four sides are the same for both pens (to form a square = maximum area), we can simplify the perimeter to 4lenghts + 4lenghts = 240 yards of fencing.
anonymous
  • anonymous
We would do well to maximize both the squares so in the end, we can have 8lengths = 240 yards.
anonymous
  • anonymous
Can you solve the length of each side from there?
anonymous
  • anonymous
30
anonymous
  • anonymous
Exactly, so each side is 30, which gives you an area of.? (A=Length*Width)
anonymous
  • anonymous
900
anonymous
  • anonymous
But I thought I had to use optimization
anonymous
  • anonymous
And since you have two pens, the max would be 900*2 sq. yards
anonymous
  • anonymous
Squares are the optimization.
anonymous
  • anonymous
so we don't have to use the derivative or anything like that?
anonymous
  • anonymous
Is this right?! Are we forgetting about the fence in the middle that separates the animals
anonymous
  • anonymous
It's two separate pens.
anonymous
  • anonymous
we are double counting the 30 side where the 2 squares are joined
anonymous
  • anonymous
Or, that is what I am getting?
anonymous
  • anonymous
two separate pens
anonymous
  • anonymous
So we should be able to get a bigger area if we share one side
anonymous
  • anonymous
I just thought we had to use the derivative of some function to get the max area
anonymous
  • anonymous
I think I solved it...if we share sides we can get a bigger area. We will need to use the derivative to solve.
anonymous
  • anonymous
Ok, that is probably right because that is what we are doing
anonymous
  • anonymous
First set up your sides as W = width. You will have 3 of these. You will also have two lengths. So you will have 2L+3W = 240. So L=120-3/2 W 120 - 3/2w _____________ | | | | w | w | w ______________ 120 - 3/2w
anonymous
  • anonymous
You with me so far? Sorry about the crappy drawing.
anonymous
  • anonymous
yea i am
anonymous
  • anonymous
Ok, that was the hard part. Now I think all you have to do is find the Area function: A = L*W or A = (120 - 3/2W) * W = \[120W - 3/2W^2 = A\] Then take the derivative: \[A \prime = 120 - 3W\] Now find where the derivative = 0 which will be it's maximum or minimum. \[0 = 120 - 3W \] From here you get the answer of W = 40. Now you can just punch it in and find the total area to b
anonymous
  • anonymous
To be 40*60 or 2400
anonymous
  • anonymous
What do you think? It looks to be right to me but double check yourself.
anonymous
  • anonymous
yea that sounds right thanks a lot!

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