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Maximizing area against a curved river

  1. Aug 12, 2008 #1
    A farmer has two miles of fence, and wants to enclose the max possible area for his cows. He will use fence on only three sides because the fourth side is a river that flows through his land. The path of the river is the parabola y=4-x^2. One of the ends of the fence is on the vertex of the parabola. and the other end of the fence is on the parabola somewhere in the first quadrant. Find the exact value of the max area.

    Here an image of how i labeled the variables on the graph and the two equations I got:

    2 = t + 2s - a

    The professor only told us to label that point (s,t) and that we'd have to differentiate the integral eventually.

    I'm confused about which letters should be treated as variables because I know I have to work with the perimeter equation to substitute it in. Any helpful hints or nudges in a good direction?
  2. jcsd
  3. Aug 12, 2008 #2


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    Welcome to PF!

    Hi Mylala88! Welcome to PF! :smile:

    I'm a little confused … could you please clarify:

    does the problem specify that the fence must go parallel to the axes (because I would expect the maximum area to be for a nearly circular fence)?

    does the problem specify that the fence is to the right of the parabola (because it's obvious that a fence to the left would enclose a larger area)? :confused:
  4. Aug 12, 2008 #3
    it's a rectangular (sort of) shaped fencing (except for the curved part). and the original picture had it the fencing on the right side.
  5. Aug 12, 2008 #4

    Forgetting (x,t)
    try this one.

    I have one equation in the diagram,
    and you should get other two

    you should get
    A = ....
    and constraint
    x^2+x+2b = 2

    Now, it's just a simple problem if you use langrange thing

    del f = lamda . del g one ...
  6. Aug 12, 2008 #5
    the only problem is that this is a calc 1 class, and we're not permitted to use that method yet. i wish i could since a friend said it'd be easier.
  7. Aug 12, 2008 #6
    Forget it, use subsitution

    initially, you have three equations and three unknowns.
    make it to one eqn one var (to area eqn)
    and differentiate the area eqn

    I don't think it matters which var you choose. You would have one var in the end ..
  8. Aug 13, 2008 #7


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    Hi Mylala88! :smile:

    Your integral [tex]\int_0^a (a\ -\ 4\ +\ x^2) dx[/tex] is wrong.

    You've taken vertical slices of thickness dx, so their length should be … ? :smile:
  9. Aug 13, 2008 #8
    Ah! I didn't even realize that! It's from 4 - t to 4. Thanks! I'll keep trying.
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