AB is a diameter of a circle and C is any point on the circle. Show that the area of Δ ABC is maximum, when it is isosceles.

Let consider AB be the width and C is any point on the circle with range r.

\[\angle ACB\text{ }=\text{ }90o\]  [angle in the semi-circle is 90o]

Let \[AC\text{ }=\text{ }x\]

Squaring on both the sides, we get

Hence, the space of \[\Delta \text{ }ABC\] is least when it is an isosceles triangle.