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Homework Help: Maxima and minima and finding the radius of the circle

  1. Mar 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Find where θ is the biggest (largest) I'll have the picture of the problem included below (pic:1)
    2. Relevant equations
    answer x= 2

    3. The attempt at a solution
    Hi, so my prefesor gave me this problem and told me to try to solve it. We already did this problem in school and got the answer that x=2.
    The trick here is that I am not allowed to use:
    the law of cosines
    therefore I have tried to circumscribe a circle and found out that the center is located at
    I would now like to know if you guys could help me calculate the perimeter or alternatively if you guys could tell me about some other way to calculate x or θ
    thank you
  2. jcsd
  3. Mar 14, 2016 #2


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    One way (depending on your trigonometric and calculus skills) would be to see that ##\theta = \theta_1 - \theta_2## with ##\tan\theta_1 = {4\over x} ## and ##\tan\theta_2 = {1\over x} ##. And, since ##0 < \theta<{\pi\over 2}##, ##\ \ \ \theta = {\rm max} ## if ##\tan\theta = \rm max##
  4. Mar 14, 2016 #3
    Thanks for the answer.
    That was the way we did it in the classroom. And the professor also mentioned that this could be solved using the law of cosine. However he asked me if I could find a way to calculate x or θ without using this two ways. I have been trying to do this for about 30 min and I am starting to doubt if this is even possible
  5. Mar 14, 2016 #4


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    You can employ the central angle theorem. But in order to do this, you have to find the correct center of the circle you have constructed on your calculation. The y coordinate can be easily seen to be 5/2, this leaves you the x coordinate of the center. Having found both the coordinate of the circle's center and its radius, you can use the central angle to do the optimization. This method requires a bit more of algebra though.
  6. Mar 14, 2016 #5
    Thanks for the reply. I really appreciate the help.
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