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Optimization Problem

  1. Nov 6, 2005 #1
    (See Attachment)
    I don't quite understand what i am supposed to optimize, and what my restriction formula is. Is QT constant? But in that case, how could i optimize PRS?
    I tried the following:
    [tex]l = PR + RS[/tex]

    [tex] PR^2 = PQ^2 + QR^2[/tex]

    [tex] cos\theta1= \frac{QR}{PR} [/tex]

    [tex] PR = \frac{QR}{cos\theta1} [/tex]

    Similarly, [tex] RS = \frac{TR}{cos\theta2} [/tex]

    So [tex] l = \frac{TR}{cos\theta2} + \frac{QR}{cos\theta1} [/tex]

    But i don't see where this could go.
    Thank you.

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    Last edited by a moderator: Nov 6, 2005
  2. jcsd
  3. Nov 7, 2005 #2


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    The problem says "show that the shortest length of the rope occurs when [itex]\theta_1= \theta_2[/tex]".

    It is the total length of the rope you wish to minimize.

    The heights of the two poles are fixed (but not given, call them "P" and "S"). The rope consists of the hypotenuses of two right triangles with angles [itex]\theta_1[/itex] and [itex]\theta_2[/itex]. You can write the total length of the rope in terms of trig functions of [itex]\theta_1[/itex] and [itex]\theta_2[/itex].

    Another way to do this, without using calculus or trigonometry, is to imagine one of the poles extending down into the ground! (This is a simple case of the "method of reflections".) Do you see that the shortest rope would be a straight line between the two pole ends? Isn't it obvious then that the two angles must be the same? The hard part is proving, geometrically, that exactly the same situation gives the shortest length for the two poles as given.
  4. Nov 7, 2005 #3
    What you said is exactly what I did. (l = total rope length)

    Can you please approve the attachment? Thank you.
  5. Nov 7, 2005 #4

    I already tried to do this problem using the suggested method. It did not work. Please give me further explanation.
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