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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.

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  1. anonymous
    • 5 years ago
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    This seems to be an optimization problem. Does the river count as one side that doesn't need to be fenced?

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    So, 2WidthCattle + 2LengthSheep + 2WidthCattle + 2 LengthSheep = 240 yards of fencing.

  4. anonymous
    • 5 years ago
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    And we know that length * width = area.

  5. anonymous
    • 5 years ago
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    Ok, Im following so far

  6. anonymous
    • 5 years ago
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    We also know that squares work the best, in terms of area.

  7. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    We would do well to maximize both the squares so in the end, we can have 8lengths = 240 yards.

  9. anonymous
    • 5 years ago
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    Can you solve the length of each side from there?

  10. anonymous
    • 5 years ago
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    30

  11. anonymous
    • 5 years ago
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    Exactly, so each side is 30, which gives you an area of.? (A=Length*Width)

  12. anonymous
    • 5 years ago
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    900

  13. anonymous
    • 5 years ago
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    But I thought I had to use optimization

  14. anonymous
    • 5 years ago
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    And since you have two pens, the max would be 900*2 sq. yards

  15. anonymous
    • 5 years ago
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    Squares are the optimization.

  16. anonymous
    • 5 years ago
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    so we don't have to use the derivative or anything like that?

  17. anonymous
    • 5 years ago
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    Is this right?! Are we forgetting about the fence in the middle that separates the animals

  18. anonymous
    • 5 years ago
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    It's two separate pens.

  19. anonymous
    • 5 years ago
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    we are double counting the 30 side where the 2 squares are joined

  20. anonymous
    • 5 years ago
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    Or, that is what I am getting?

  21. anonymous
    • 5 years ago
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    two separate pens

  22. anonymous
    • 5 years ago
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    So we should be able to get a bigger area if we share one side

  23. anonymous
    • 5 years ago
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    I just thought we had to use the derivative of some function to get the max area

  24. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Ok, that is probably right because that is what we are doing

  26. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    You with me so far? Sorry about the crappy drawing.

  28. anonymous
    • 5 years ago
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    yea i am

  29. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    To be 40*60 or 2400

  31. anonymous
    • 5 years ago
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    What do you think? It looks to be right to me but double check yourself.

  32. anonymous
    • 5 years ago
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    yea that sounds right thanks a lot!

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