What is the maximum perimeter for a rectangle inside an ellipse?

In summary, the problem involves finding the longest perimeter possible for a rectangle placed symmetrically inside an ellipse defined by the equation x^{2} + 4y^{2} = 1. The perimeter of the rectangle is given by P = 2x + 2y and the problem can be solved by extremising the function f(x,y) = 2x + 2y on the ellipse. This involves calculating partial derivatives and looking for consistent solutions, but the exact method for solving this is not provided in the given conversation.
  • #1
Hart
169
0

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:
 
Physics news on Phys.org
  • #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.
 
  • #3
? :confused: ?
 
  • #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.
 
  • #5
Hart said:
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
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:

What is a rectangle inside an ellipse?

A rectangle inside an ellipse is a geometric figure where a rectangle is completely contained within an ellipse. The edges of the rectangle touch the edges of the ellipse at four points, creating four curved corners.

What are the properties of a rectangle inside an ellipse?

The properties of a rectangle inside an ellipse include: the length of the sides of the rectangle are parallel to the major and minor axes of the ellipse, the diagonals of the rectangle are equal in length and bisect each other at the center of the ellipse, and the area of the rectangle is equal to the area of the ellipse.

How is a rectangle inside an ellipse different from a regular rectangle?

A rectangle inside an ellipse is different from a regular rectangle in that its sides are curved instead of straight. This means that the angles of the rectangle are not 90 degrees, and the length of each side may vary depending on its position within the ellipse.

What is the equation for finding the area of a rectangle inside an ellipse?

The equation for finding the area of a rectangle inside an ellipse is A = ab, where a and b are the lengths of the semi-major and semi-minor axes of the ellipse, respectively. This is the same equation used to find the area of the ellipse itself.

How is a rectangle inside an ellipse used in real life?

A rectangle inside an ellipse can be found in many real-life applications, such as architectural designs, engineering structures, and even in art and design. For example, the shape of a television or computer screen can be represented by a rectangle inside an ellipse to ensure that the image is displayed evenly and without distortion.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
842
  • Advanced Physics Homework Help
Replies
13
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
933
  • Advanced Physics Homework Help
Replies
9
Views
864
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
975
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
370
Back
Top