Find max area of rectangle inscribed in triangle

Discussion in 'Undergraduate Math' started by lucky_luke, Apr 15, 2006.

  1. lucky_luke

    lucky_luke Guest

    hi

    I don't know how to even begin.

    I have to find maximum area of rectangle inscribed inside of rectangle with height H and hypotenuse c.


    In right triangle with hypotenuse c = 8 and angle = 30 is inscribed rectangle with maximum possible area.
    Find sides a and b of triangle, if two sides of rectangle
    lie on a and b

    thank you
     
    lucky_luke, Apr 15, 2006
    #1
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  2. The way to begin, if you expect to get help, is to state the problem
    exactly as it is given to you instead of adding your confusion to the
    statement of the problem.

    --Lynn
     
    [Mr.] Lynn Kurtz, Apr 15, 2006
    #2
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  3. lucky_luke

    Jasen Betts Guest

    i'll assume the second "rectangle" is a triangle.

    Half the area of the triangle. Interestingly the converse is true too.

    Bye.
    Jasen
     
    Jasen Betts, Apr 15, 2006
    #3
  4. lucky_luke

    G.E. Ivey Guest

    hi
    ?? You posted several different problems. I assume you mean: given a right triangle find the area of the largest rectangle that can be inscribed in the right triangle with the right angle of the triangle as one corner of the rectangle.

    "Find sides a and b of triangle, if two sides of
    rectangle lie on a and b" makes no sense at all. I THINK you are saying you have a right triangle with hypotenuse c = 8 and angle = 30 degrees.

    Then the first thing you need to do (the rectangle is irrelevant at this point) is find the lengths of sides a and b: imagine the triangle reflected about the side adjacent to the 30 degree angle to make a triangle with angle 60 degrees. It's easy to see that that is an equilateral triangle with side of length 8. Side a (opposite the 30 degree angle) of the right triangle is half the third side of the equilateral triangle and so has length 4. By the Pythagorean theorem, side b, has length sqrt(64- 16)= sqrt(48)= sqrt(16*3)= 4sqrt(3).

    Now, for the rectangle. Here's how I would do it: Set up a coordinate system with x axis along a and y axis along b. One corner of the rectangle is at (0,0), one corner is on the x-axis, one corner is on the y-axis and one corner is on the hypotenuse. The hypotenuse is a line passing through (4, 0) and (0, 4sqrt(3)) and so x/4+ y/(4sqrt(3))= 1. Solving for y: y= -sqrt(3)x+ 4sqrt(3). That is, if the the vertex on the x-axis is at (x,0) then the vertex on the hypotenuse is at (x, 4sqrt(3)-sqrt(3)x) and so the vertex on the y-axis is at (0, 4sqrt(3)- sqrt(3)x). The area of the rectangle is the product of those lengths: A= 4sqrt(3)x- sqrt(3)x^2= (-sqrt(3)(x^2- 4x).
    One method of finding maximum area is to take the derivative of that but for a quadratic, we can just complete the square:
    A= (-sqrt(3))(x^2- 4x+ 4- 4)= (-sqrt(3))((x- 2)^2-4).
    (x-2)^2 is never negative so (x-2)^2- 4 is larger than -4 as long as x is not 2 and has minimum value (-4) at x= 2. That is, A is less than 4sqrt(3) as long as x is not 2 and has maximum value 4sqrt(3) when x= 2.

    That is interesting- the triangle has base of length 4 and height of length 4sqrt(3) and so area (1/2)(4)(4sqrt(3)= 8sqrt(3). As Jasen Betts said, the area of the largest rectangle inscribed in this way is half the area of the triangle itself.
     
    G.E. Ivey, Apr 15, 2006
    #4
  5. lucky_luke

    Virgil Guest

    This maximum of "half the area" is true for an arbitrary triangle.
    One such rectangle of maximal area may always be formed with one side of
    the rectangle along a longest side of the triangle and the opposite side
    of the rectangle having vertices at midpoints of the other two sides of
    the triangle.
     
    Virgil, Apr 15, 2006
    #5
  6. lucky_luke

    lucky_luke Guest

    I really did screw up my post and I'm sorry for that.
    Here is original question reposted and "repaired"

    How would I solve first this,since in that example
    vertices of rectangle may very well be located only
    on two sides of triangle( if triangle has sides a, b
    and c, then two vertices could be located on c,
    and third vertex on either a or b,and 4. vertex would
    be located somewhere inside a triangle not touching
    any of the three sides )?Of course, this would happened
    only if one of the angles inside triangle would bigger
    that Pi/2.
     
    lucky_luke, Apr 15, 2006
    #6
  7. lucky_luke

    Virgil Guest

    If the given triangle is to have a hypotenuse, it must be a right
    triangle, and one can then choose arbitrarily to have one side of a
    maximal rectangle along any one side of the triangle. And if one does
    not choose to have the rectangle along the hypoteneuse, two sides of the
    maximal rectangle will have to lie along the shorter two sides of the
    triangle.

    It is only in triangles with one angle greater than a right angle that
    the maximal rectangle is limited to lying along only one side, the side
    opposite the greatest angle.
     
    Virgil, Apr 15, 2006
    #7
  8. lucky_luke

    lucky_luke Guest

    I really did screw up my post and I'm sorry for that.
    It was bad choice of words on my part. I shouldn't
    use the word hypotenuse since problem didn't
    specify it. I don't think we are dealing with right
    triangle. Any idea how I would solve it?
     
    lucky_luke, Apr 16, 2006
    #8
  9. Consider putting one side of your rectangle along a longest side of the
    triangle with its other vertices on the other two sides and use
    properties of similar triangles.
     
    Virgil Hancher, Apr 17, 2006
    #9
  10. lucky_luke

    lucky_luke Guest

    Can someone show me?
     
    lucky_luke, Apr 17, 2006
    #10
  11. lucky_luke

    Jasen Betts Guest

    you can divide any triangle into two right triangles and the maximal
    rectangles for both will gave the same side length on thee shared part of
    the triangle.

    the answer's the same; the area of the rectangle will be half that of the
    triangle.
     
    Jasen Betts, Apr 18, 2006
    #11
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