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Minima of Hypotenuse

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data
    a and b are two perpendiculars drawn to the sides from a point on the hypotenuse of a right angled, isoceles triangle.

    Find the minimum value of the hypotenuse in terms of a and b.

    2. Relevant equations
    ABC is a right angled, isoceles triangle, with AC, the hypotenuse, h.

    X is a point on AC, from which the perpendiculars a and b are drawn to the sides AB and BC respectively.

    AB = BC= h / (sq.root 2)


    3. The attempt at a solution
    Tried to represent h in terms of a single variable (some angle theta and trigonometric ratios of theta). This would eliminate two variables a and b. With this, h can be differentiated and equated to zero for Minima, and confirming it with 2nd derivative. But could not manage eliminating a and b and converting to a single variable equation.
     

    Attached Files:

  2. jcsd
  3. Jan 25, 2009 #2

    Mark44

    Staff: Mentor

    Since this is an isosceles right triangle, theta = pi/4, which it doesn't seem you have used. I don't know whether this is a helpful clue.
     
  4. Jan 25, 2009 #3
    Hi! Thanks for your time. I was not referring to theta as angle BAC or BCA, which is pi/4. For a point X, a and b are fixed, and are related (except that I am unable to figure it out straight). If I can represent h as an equation of a alone, or b alone, or some other variable theta on which a and b are dependent, then I can differentiate h and equate to 0 for Minima. This is where I am stuck. Or, is there some other way?
     
  5. Jan 26, 2009 #4
    Why is the attachment pending approval?
    why is my question being unanswered?
    Believe me i tried my best before posting up the question here.
    Somebody please help me?
     
  6. Jan 26, 2009 #5

    gabbagabbahey

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    Homework Helper
    Gold Member

    All attachments are screened by site admin before they are approved to be hosted at PF. This sometimes takes days, so it is quicker for you to host the image file at a free hosting site like imageshack.us and simply post a link to the image.

    Until we can view your attached file, we have no way of knowing where your mistakes might be.
     
  7. Jan 26, 2009 #6
  8. Jan 26, 2009 #7

    gabbagabbahey

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    Gold Member

    This question doesn't make sense to me; the length of the hypotenuse doesn't actually depend on 'a' and 'b', so it can't be minimized with respect to those variables.

    Are you sure the question isn't "Find the minimum length of the line segment AX in terms of a and b"?

    Is there more to the question than what you have posted here?
     
  9. Jan 27, 2009 #8
    Yes you're right. I went to school today and got the question sorted out, it wasn't isosceles. A friend of mine verbally stated out the question he had picked up from some guide book, and I seem to have taken it down wrongly, mixed up multiple questions!

    Sorry about that. I panicked as I had an exam today!

    The correct question is:

    A point on the hypotenuse is at a distance a and b from the sides of the triangle. show that the maximum length of hypotenuse is (a^(2/3) +b^(2/3))^(3/2).

    btw, i had this question on the exam and i solved it correctly.

    I sincerely apologize for this mistake once again.
    Thanks for your help.
     
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