This can be solved as a standard calculus maximization problem, or without any calculus as follows.  Refer to the diagram on the left.  

Rotating the hexagon clockwise will decrease the length of the diameter BB', while rotating it counterclockwise will decrease AA', therefore, this must be the orientation of the inscribed regular hexagon of largest area.  Let m be the length of one side of the hexagon. It's easy to compute that the area of the hexagon is then 3√3m²/2.  To determine m, start with the observations that x = mcos(45^◦), y = msin(15^◦), and z = mcos(15^◦), then use the fact that x + y + z = 1 to conclude that m = 1/(cos(45^◦) + sin(15^◦) + cos(15^◦)).  This gives an area of 3√3 - 9/2 ≈ .6961. 

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ã2004 Alberto L. Delgado