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Optimization Using Differentiation

  1. Nov 11, 2003 #1
    [SOLVED] Optimization Using Differentiation

    I have an assignment in which we are to optimize problems using a given 6-step process. More or less it involves Max/Min differentiation.

    On of the problems are as follow;
    Enclosing the Largest Area
    The owner of the Rancho Los Feliz has 3000 yd of fencing material to enclose the rectangular piece of grazing land along the striaght portion of a river. If fencing is not required along the river, what are the dimensions of the lagrgest area that the he can enclose? What is the area?

    I under stand that...

    i understand one of the sides can be added to the other 3 sides, however, i'm not sure how to make this a function.
  2. jcsd
  3. Nov 11, 2003 #2
    Have you learned how to use Lagrange multipliers to handle constraints?
  4. Nov 11, 2003 #3


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    Re: Re: Optimization Using Differentiation

    Try figuring out the area of the enclosure based on the lenght of one of the sides:
    A(x)=blah blah blah.

    you should be able to construct a function like that.

    P.S. The farmer doesn't have to put fence where the river is, so you should probably use:
  5. Nov 12, 2003 #4


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    What you have to do is substitute one into the other.
    a = xy
    p = x + 2y

    lets rearrange the perimeter equation in terms of x
    x = p - 2y
    and since we know p, we can can a bit farther
    x = 3000 - 2y

    since x = 3000 - 2y
    a = (3000 - 2y)y
    a = 3000y - 2y^2

    since the area changes when we change the y, lets find when our differential
    da/dy = 3000 - 4y
    since the maximum area is when the area stops increasing, we equate to 0
    0 = 3000 - 4y
    after solving, we get
    y = 750

    now fill that back into our original equation of x = 3000 - 2y
    x = 3000 - 2(750)
    now solve
    x = 1500

    now fill in for the area formula, a = xy
    a = 1500 * 750
    a = 1125000

    Right on :smile:
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