Consider a given square with edges of length . Prove that the inscribed square with minimum area has edges of length .
(Note: The picture is of an arbitrary inscribed square to illustrate what and are. It is definitely not the particular inscribed square with minimal area that we are looking for.)
Proof. We start with a square with edges of length . Let where and are the lengths of the two sections of created by the point at which the corner of the inscribed square meets the edge of the outer square. Let denote the length of the edge of the inscribed square. Then, implies . So,
Let this be our function . Then we take the derivative,
From this we have when and
Therefore, is decreasing when and increasing when . Hence, has a minimum at . Using our equation for , we have
Finally, solving for the length of the edge ,