[AntonioLao]

In the history of the calculus of variations, famous problem of Princess Dido was the first recorded event applying isoperimetric solution that maximizes areas of closed plane curves. One version of her story is the following: She was unhappy and dissatisfied being a princess in the oppressive city of Tyre in ancient Phoenicia (now southern Syria). Thereafter she ran away from home to dwell on the coast of northern Africa, along the Mediterranean and was forced to purchase land that could be encompassed by a large piece of cowhide. She proceeded to cut the hide into very thin string-like strips and tied end to end. Moreover, she chose land already bounded by the inland sea on one side so that the shoreline spares lying of more needed hide. Her maximized layout is that of a semi-circle with the straightest shoreline being the diameter. On the other hand, if the sides are to be perpendicular line segments then the maximum layout would have to be a square with sides each ¼ the lengths of the given fixed perimeter hide.

The square is really a 4-sided regular polygon and the general formula for finding the area is ½ the product of the apothem and the perimeter. For an n-gon where n (number of sides) approaches infinity, the apothem approaches the radius of the circumscribed circle if and only if the perimeter of the polygon is set equal to the circumference.

For an ellipse with the same perimeter the exact formula requires knowledge of elliptic functions and their integrals. Thus a solution delves into the realm of complex analysis. Since an elliptic function is single-valued but doubly periodic with one singularity at infinity. However, starting with an approximate formula, the perimeter is the product of 2p and the square root of half the sum of the squares of the semi-major axis (a) and semi-minor axis (b). But the area is just the product of pand these axes.

If the perimeter is set to unity then the radius of a circle with unit circumference is 1/2p and its area is 1/4p. An ellipse with equivalent perimeter and area gives a polynomial equation in terms of a and b as a²-2ab+b²=(a-b)²=0. As the beginning for a principle of least action, these two degenerate roots imply that for maximum area the difference of the semi-axes must approach zero.