This follows immediately from the relation between the square side x and the axes of the ellipse inscribed in the square. They satisfy the relation (prove it) a² + b² = x²/2. Under this condition, for constant x, the area of the ellipse E(a,b) = pi*a*b, becomes a maximum when a = b = x/2.

For an application of this fact to the problem of finding the maximal ellipse, inscribed in an arbitrary parallelogram, look at the file ParaCircumscribed.html .

For a similar discussion concerning the minimal in area ellipse, circumscribed on a square , look at the file MinimalEllipse.html .

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