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Optimization of the Area of a Triangle

  1. Oct 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 4. Solve by writing the area as a function of θ
    upload_2014-10-6_15-43-49.png
    2. Relevant equations
    A=1/2 (bh)

    3. The attempt at a solution

    Given the side h and the hypotenuse 4, we can find the base of the smallest triangle to be sqrt(16-h2) which means that the base of the largest triangle is 2(sqrt(16-h2)). Its height can be considered as h+4, which would give a side length of sqrt(8h+32). So the area would be:
    sqrt(16-h2)(h+4). However, they want it to be solved in terms of theta, which is the part that I am not sure of.
     
  2. jcsd
  3. Oct 6, 2014 #2

    BiGyElLoWhAt

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    can you express h in terms of theta? express the sides of the triangle in terms of theta.
     
  4. Oct 6, 2014 #3
    for convenience, I'll say c= sqrt(8h+32). Would it be correct if I said the height was c(sinθ) and the base of the largest triangle 2c(cosθ)?
     
  5. Oct 6, 2014 #4

    LCKurtz

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    I don't know where you got ##c = \sqrt{8h+32}##. It would help if you labeled your picture. If you note that the angle between the two sides that you have labeled ##4## and ##h## is ##2\theta## that will help you get the two legs of that little right triangle in terms of ##\theta##.
     
  6. Oct 6, 2014 #5
    How did you determine that angle to be 2θ?
    And to answer your question, I used the pythagorean theorem to find one of the side lengths
    upload_2014-10-6_17-20-23.png
     
  7. Oct 6, 2014 #6

    RUber

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    From what you have above, it seems that the height ##(h+4)= Ccos(\theta)## Where, as you mentioned, C is the length of the hypotenuse. You should be able to solve that for h in terms of theta and plug that into the formula for area.
     
  8. Oct 6, 2014 #7
    If I solve for h, that would give me Ccosθ-4=h. Is it ok to have an extra variable (C) in there?
     

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  9. Oct 6, 2014 #8

    LCKurtz

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    That upper triangle is an isosceles triangle with equal sides of ##4##. Its angles are ##\theta,\theta, 180 -2\theta##. Label them. The supplement of that last one is the angle you are looking for.

    Once you have that you have both ##h## and the other leg in terms of ##\theta##.
     
  10. Oct 6, 2014 #9
    Are these labeled correctly?
    upload_2014-10-6_20-53-53.png
    If those are correct, can I use the law of sines such that

    h/(sin90-2θ) = 4/sin90
    h = 4(sin90-2θ)
    h=4(1-2θ)
    h=4-8θ
     

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  11. Oct 6, 2014 #10

    LCKurtz

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    You mean ##h/\sin(90-2\theta)##. Parentheses are important.

    Which is why you then get nonsense. Also, it is silly to use the law of sines given the picture. You have adjacent sides to the ##h## and ##4## for the angle ##2\theta##. What trig function does that give you for ##h##? Similarly for the other leg.
     
  12. Oct 6, 2014 #11
    That would be cos2θ=h/4
    4cos2θ=h
     
  13. Oct 6, 2014 #12

    LCKurtz

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    Yes. And what about the other leg?
     
  14. Oct 6, 2014 #13
    Well since h=4cos2θ, couldn't we just substitute it into the expression to get sqrt(16-(4cos2θ)^2) for the base?
     
  15. Oct 6, 2014 #14

    RUber

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    That looks a lot like ##4(\sqrt(1-\cos^2 2\theta)##. It seems like the long way to get to the opposite side from the angle ##2\theta## with hypotenuse 4.
     
  16. Oct 6, 2014 #15

    LCKurtz

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    Yes, you could, but you would want to simplify it. But, again, it is silly to do it that way. What trig formula uses opposite/hypotenuse?
     
  17. Oct 6, 2014 #16
    That would be sine. So if I were to use that, I would get sin(2θ)= [sqrt(16-(h^2))]/4. But I don't understand what this would accomplish since we're trying to look for max area, and that involves base and height
     
  18. Oct 6, 2014 #17
    Ahhh never mind. I think I see it now
     
  19. Oct 6, 2014 #18

    LCKurtz

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    So instead of calling that leg ##\sqrt{16-h^2}## you can call it ##4\sin(2\theta)##. Now you have both legs of that little triangle in terms of ##\theta##. Now you can express the area of the original big triangle in terms of ##\theta##. Then you are ready for the calculus problem of maximizing the area. Too bad the trig is holding you back so badly. Perhaps you should review it.
     
  20. Oct 6, 2014 #19
    So A= 1/2 (4cos(2θ))(4sin(2θ))
    = 8cos(2θ)sin(2θ)

    deriving this, we get A'= -16(sin(2θ))2 + 16(cos(2θ))2

    setting it to zero, we get θ= 30 degrees

    substituting this for θ we get 2(sqrt(3))

    However, another part of the question was solving the area in terms of h (I used A= [sqrt(16-h2)](h+4)

    Following the same steps as I did for the one above, I get h=-4 and 2. h can't be -4 because if we use it, h+4 would go to zero, and the whole area would be zero. So h can only be 2.

    Substituting this for h I get 12(sqrt(3))

    What am I doing wrong?
     
  21. Oct 6, 2014 #20

    LCKurtz

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    Isn't the height of your triangle ##4+4\cos(2\theta)##? And why the 1/2 out in front?
     
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