- #1

Hart

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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