16: The condition that the rectangle is inscribed in the half-circle means that two verteces must be on the diameter and two on the half-circle itself. Connecting one such vertex with the center of the circle one gets a useful triangle.

The resulting equation

(x/2)² + y² = R²

The function

should be maximized over the set M = [0,2R] (the longest side of a rectangle in a half-circle cannot be longer than the diameter). Use the appropriate algorithm.

Actually, an experienced problem solver would maximize the function g(x) = 2f (x) in order to get rid of constants that do not influence the outcome.

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