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Optimisation problem

  1. May 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A right angled triangle with sides of length a, b and 2r is trapped within a semi-circle of fixed radius r. Given that all three vertices are on the semi-circle, find the values of a and b that will maximise the area of the triangle

    http://img100.imageshack.us/img100/1804/areaoftrainglebl3.th.jpg [Broken]

    2. Relevant equations
    A(circle) = pi r ^2
    A(triangle) using hero's formula (sqrt(s(s-a)(s-b)(s-c))) where s=(a+b+c)/2
    A(triangle)= (b x h)/2

    3. The attempt at a solution
    I am familiar with optimisation skills but am really stumped with this question, given that it is a scalene triangle.
    If someone could give me a pointer in the right direction as to what formulas to use or attempt to derive, it would be appreciated
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 26, 2007 #2
  4. May 26, 2007 #3
    Hint - Since the vertices are always on the semi-circle, then the triangle will always be a right-triangle -- area of triangle will be (1/2) a * b
  5. May 26, 2007 #4

    that is so obvious, cant believe i didn't figure that out.
    I suppose I just didn't visualise the traingle in that way OR notice that it actually told me that it was right-angled ... :blushing:

    i shall now attempt to optimise it
    thanks fellas
  6. May 26, 2007 #5

    so i have found that the area of the triangle is (ab)/2
    Now i want to optimise this so that the area is maximised, so we need another equation.

    I know that the area of the half-circle is (pi x r^2)
    how can i now derive some other equation to maximise a and b?
  7. May 26, 2007 #6
    There is a relation - What can you say about the sides of a right triangle?
  8. May 26, 2007 #7
    pythagoras ....
    a^2 + b^2 = (2r)^2

    take that in terms of 'a' or 'b' and then replace than in the area equation, find the derivate and then optain max/min .... cheers

    sorry not thinking very well tonight ... thankyou for you patience.
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