## anonymous 5 years ago 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.

1. anonymous

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

2. 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.

3. anonymous

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

4. anonymous

And we know that length * width = area.

5. anonymous

Ok, Im following so far

6. anonymous

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

7. 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.

8. anonymous

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

9. anonymous

Can you solve the length of each side from there?

10. anonymous

30

11. anonymous

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

12. anonymous

900

13. anonymous

But I thought I had to use optimization

14. anonymous

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

15. anonymous

Squares are the optimization.

16. anonymous

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

17. anonymous

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

18. anonymous

It's two separate pens.

19. anonymous

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

20. anonymous

Or, that is what I am getting?

21. anonymous

two separate pens

22. anonymous

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

23. anonymous

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

24. 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.

25. anonymous

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

26. 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

27. anonymous

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

28. anonymous

yea i am

29. 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

30. anonymous

To be 40*60 or 2400

31. anonymous

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

32. anonymous

yea that sounds right thanks a lot!

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