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Rectangle in a circle

  1. May 3, 2005 #1
    The problem asks me to show that the maximum possible area for a rectangle
    inscribed in a circle of radius R is 2R^2. It also gives a hint saying that I should first maximize the square of the area.

    I set the problem up as xy = 2R^2. I decided to work on the left side and wrote it as x(2R). But, I don't know where to go from here. Thanks.
  2. jcsd
  3. May 3, 2005 #2


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    You need to start up with a relationship such like

    [tex] x^2 + y^2 = (2R)^2 [/tex]

    and like you noted

    [tex] xy = Area[/tex]

    The use the concept of derivate to maximize for the area.

    The standard procedure is to find a or more relations F(x,y) and then turn it into a function F(x), so you can use the concept of derivative.

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    Last edited: May 3, 2005
  4. May 21, 2005 #3
    If You know some trigonometry i.e. 2 sinA cosA = sin2A.

    Solution : Since the rectangle is inscribed in a circle hence, its diagonal,D, is equal to the diamter,2R, of the given circle. i.e. D = 2R.
    Now consider the triangle formed by the diagonal and two adjacent sides of our rectangle. It is a right triangle. Hence its sides can be expressed as D(diagonal the hypotenuse) , D*sinA and D*cosA(the perpendicular and base ).
    Note : they also satisfy the pythagoras theorem D^2 = (D*sinA)^2 + (D*cosA)^2.

    hence area of rectangle = product of adjacent sides
    = DsinA * DcosA
    = D*D* (sinA*cosA)
    = 2R*2R*(sinA cosA)
    = 2(R^2)*(sin 2A )
    now R is fixed for a given circle. and A may vary.
    for maximum area, sin2A must be maximum. which is 1
    hence maximum area is 2R^2.
    Note : sin 2A is 1 only if A =45 degree
    which means that sinA = cos A
    hence adjacent sides of the rectangle D*sinA and D*cosA become equal.
    This justifies that in order to maximise its area the rectangle must be made into a square.
  5. May 21, 2005 #4


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    I like the geometrical solution proposed in post #3,but,to carry on with the idea in post #2,i'd say the simplest & most elegant way is to use a Lagrange multiplier.

  6. May 21, 2005 #5


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    this is a trivial calculus problem, to maximize A= xy when x^2 + y^2 = 4R^2, as x runs from 0 to 2R.

    If we take the hint and maximize A^2 = x^2 y^2 = x^2 [ 4R^2 - x^2]

    = 4R^2 x^2 - x^4, the derivative is 8R^2 x - 4x^3, which is zero when x = 0 and x = sqrt(2)R.

    the max is obviously not at x= 0 hence occurs at x = sqrt(2)R (since on a closed interval there must be a max and it must occur at a critical point.
    Last edited: May 21, 2005
  7. May 21, 2005 #6


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    the solution is also obvious geometrically as follows: picture the level curves of the function xy, i.e. the hyperbolas xy = c. if you know what these look like, it is obvious that they are symmetrical about the line x=y, and the point on each hyperbola which is nearest the origin lies on this line.

    Hence if we assume also that (x,y) lies on a circle of given radius, then the hyperbola xy = c which meets the circle and has largest value of c, is the one tangent to the circle at the point (sqrtc,sqrtc).
    Last edited: May 21, 2005
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