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Applied minimum problem

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the dimensions of an isosceles triangle of least area that can be circumscribed about a circle of radius R.


    2. Relevant equations
    Let a,b, and c denote the lengths of the sides of this triangle and c=b. Then the angles opposite these sides are A, B and C respectively, (B=C). Let B' be the angle between the radius R and the side a. See attached figure.


    3. The attempt at a solution
    The quantity to be minimumized is the area of the triangle. I am not too sure how to go about this but the area of the triangle [tex] S=\frac{a}{2}(R+R \sin(B'))[/tex]. Now 2B'+2A=pi, therefore [tex] B' =\frac{\pi}{2}-A \mbox{ therefore} \sin (B') = \sin(\frac{\pi}{2}-A) = \sin(\frac{\pi}{2}+A)[/tex]. Therefore [tex]S=\frac{a}{2}R(1+\sin(\frac{\pi}{2}+A))[/tex].
    If I differentiate this implicitly I get [tex] \frac{a}{2}+\frac{a}{2}R\frac{d}{dR}R \sin(\frac{\pi}{2}+A) + \frac{a}{2} \sin(\frac{\pi}{2}+A)=0[/tex].
    I need to find an expression for the [tex] \sin(\frac{\pi}{2}+A)[/tex]. I cannot see one suitable.
    thanks for helping.
     

    Attached Files:

  2. jcsd
  3. Apr 17, 2010 #2

    LCKurtz

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    I notice that only 5 people (6 counting me) have viewed your document. Most people will not go to the trouble of opening a Word document to see a simple figure. Had more people looked, you might have learned sooner that you are trying to work the wrong problem. The question asks for a triangle circumscribed about the circle, not inscribed as you have pictured.

    If you think about it the minimum area inscribed such "triangle" would have area 0.
     
  4. Apr 18, 2010 #3
    Thanks for the reply. I hope the new word attachment has the situtation described correctly. I still have the problem of finding the minimized area of this triangle though. The area of the triangle is [tex] T = \frac{a}{2} b \sin(C) [/tex] I have a,b and sinC to express in terms of the radius of the circle R. or what should my approach be? Maybe I should differentiate implicitly the area, T, first, and then substitute in expressions for a, b, sinC. Finding suitable expressions is the problem for me. Thanks in advance.
     
  5. Apr 18, 2010 #4
    I do not know why this new attachment didn't post in my last post.
     

    Attached Files:

  6. Apr 19, 2010 #5
    [tex] T=\frac{a}{2} c \sin(B) \mbox{therefore we have } T = \frac{a}{2} c \frac{x+2R}{c} [/tex] The c's cancel out, now the problem is to find an expression for a from the figure in terms of x and R only? Or is there a more simple way to do this?
     
  7. Apr 20, 2010 #6
    Ok, I found another formula for the area of this triangle area =R*s, where R =radius of the circle just touching each of the sides and lies inside the triangle, and [tex] s = \frac{a+b+c}{2} [/tex], but if you differentiate this w.r.t., R and put result equal to 0, you get s=0 i.e., [tex] \frac{a+2b}{2}=0 [/tex] Then a=-2b clearly wrong. Please help and Thanks very much for the help.
     
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