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.

- anonymous

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- anonymous

This seems to be an optimization problem. Does the river count as one side that doesn't need to be fenced?

- 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

So, 2WidthCattle + 2LengthSheep + 2WidthCattle + 2 LengthSheep = 240 yards of fencing.

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- anonymous

And we know that length * width = area.

- anonymous

Ok, Im following so far

- anonymous

We also know that squares work the best, in terms of area.

- 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

We would do well to maximize both the squares so in the end, we can have 8lengths = 240 yards.

- anonymous

Can you solve the length of each side from there?

- anonymous

30

- anonymous

Exactly, so each side is 30, which gives you an area of.? (A=Length*Width)

- anonymous

900

- anonymous

But I thought I had to use optimization

- anonymous

And since you have two pens, the max would be 900*2 sq. yards

- anonymous

Squares are the optimization.

- anonymous

so we don't have to use the derivative or anything like that?

- anonymous

Is this right?! Are we forgetting about the fence in the middle that separates
the animals

- anonymous

It's two separate pens.

- anonymous

we are double counting the 30 side where the 2 squares are joined

- anonymous

Or, that is what I am getting?

- anonymous

two separate pens

- anonymous

So we should be able to get a bigger area if we share one side

- anonymous

I just thought we had to use the derivative of some function to get the max area

- 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

Ok, that is probably right because that is what we are doing

- 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

You with me so far? Sorry about the crappy drawing.

- anonymous

yea i am

- 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

To be 40*60 or 2400

- anonymous

What do you think? It looks to be right to me but double check yourself.

- anonymous

yea that sounds right
thanks a lot!

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