Consider all rectangles of a fixed area. Prove that the square admits the smallest circumscribed circle.
Proof. Let and denote the lengths of the sides of the rectangle, the radius of the circumscribed circle, and the area of the square. (This is similar to the previous exercise, except this time the area of the square is fixed instead of the circle being fixed.) Then,
Furthermore, since we have a right triangle whose hypotenuse is the diameter of the circle (therefore is ) and whose legs are the sides of the square we have,
We then want to find the minimum value of this function (since this function gives us the radius of the circle). Calling the function and taking its derivative we have,
Then, setting this equal to zero,
This critical point is a minimum since
Thus, the radius of the circle is minimized when , so the rectangle is a square