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Rectangle Inside An Ellipse

  1. Mar 12, 2010 #1
    1. The problem statement, all variables and given/known data

    A rectangle is placed symmetrically inside an ellipse (i.e. with all four corners
    touching the ellipse) which is defined by:

    [tex]x^{2} + 4y^{2} = 1[/tex]

    Find the length of the longest perimeter possible for such a rectangle.

    2. Relevant equations

    Within the problem statement and solutions.

    3. The attempt at a solution

    Firstly rearranged the given equation:

    [tex]x^{2} + 4y^{2} = 1 \implies x^{2} + 4y^{2} - 1 = 0[/tex]

    Then stated the equation for the perimeter of the rectangle:

    [tex]P = 2x + 2y[/tex]

    Hence need to extremise:

    [tex]f(x,y) = 2x + 2y[/tex]

    .. on the ellipse:

    [tex]g(x,y) = x^{2} + 4y^{2} - 1 = 0[/tex]

    Therefore:

    [tex]F(x,y,\lambda) = f + \lambda g = 2x + 2y + \lambda\left(x^{2} + 4y^{2} - 1\right)[/tex]

    Then calculate partial derivatives:

    [tex]\frac{\partial F}{\partial x} = 2 + 2y + \lambda\left(x^{2} + 4y^{2} - 1\right) = 0[/tex]

    [tex]\frac{\partial F}{\partial y} = 2 + 2x + \lambda\left(x^{2} + 4y^{2} - 1\right) = 0[/tex]

    [tex]\frac{\partial F}{\partial \lambda} = 2x + 2y + \left(x^{2} + 4y^{2} - 1\right) = 0[/tex]

    Now I need to look for 'consistent solutions' (i.e. values for x, y, \lambda) within those equations, but I'm a bit stuck with that now :frown:
     
  2. jcsd
  3. Mar 12, 2010 #2
    How about you parameterize only one corner of the rectangle? It will be sufficient to define the whole rectangle and you only meed to move the point along the ellipse.
     
  4. Mar 12, 2010 #3
    ? :confused: ?
     
  5. Mar 12, 2010 #4
    The equation for (half an) ellipse is [tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex], where a and b are the lengths of the axes and given in the problem, but I will generalize here. Expressing y in terms of x shouldn't be a problem.

    From here, assign, say, the upper right corner an x value which is bounded to [0,a]. Get the y-value from the expression above and you have the coordinates of your first corner of the rectangle. Per restriction of symmetry the sides of the rectangle will be parallell to the coordinate axes. You now have enough information to obtain the perimeter of the rectangle. Optimize.
     
  6. Mar 12, 2010 #5


    I think there's a calculation error at partial differentiation..
    [tex]\frac{\partial F}{\partial x} [/tex]
    should be [tex]2 + \lambda\left(2x\right)[/tex]
     
  7. Mar 12, 2010 #6
    Thought I may have made a mistake with calculations of the partial derivatives :frown:

    So, should be then:

    [tex]\frac{\partial F}{\partial x} = 2 + 2x\lambda = 0[/tex]

    [tex]\frac{\partial F}{\partial x} = 2 + 2y\lambda = 0[/tex]

    [tex]\frac{\partial F}{\partial x} = \left(x^{2} + 4y^{2} - 1\right) = 0[/tex]

    .. correct?
     
    Last edited: Mar 13, 2010
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