This follows immediately from the relation between the square side x and the axes of the ellipse circumscribed on the square. They satisfy the relation (prove it) (1/a^2 + 1/b^2 = 4/x^2). Under this condition, for constant x, the area of the ellipse E(a,b) = pi*a*b, becomes minimum when a = b = x/sqrt(2).

For an application of this fact to the problem of finding the minimal ellipse, circumscribed on an arbitrary parallelogram, look at the file ParaCircumscribed.html .

For a similar discussion concerning the maximal in area ellipse, inscribed in a square , look at the file MaximalEllipse.html .

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