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Rectangle Inside An Ellipse

  • Thread starter Hart
  • Start date
  • #1
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Homework Statement



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.

Homework Equations



Within the problem statement and solutions.

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:
 

Answers and Replies

  • #2
834
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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.
 
  • #3
169
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? :confused: ?
 
  • #4
834
1
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.
 
  • #5
1
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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:


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]
 
  • #6
169
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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:

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