Given: a circle with AB as diameter and C is any point on the circle

To show: area of Δ ABC is maximum, when it is isosceles

Explanation: Let r be the radius of the circle, then the diameter will be

AB = 2r

Now we know, any angle in the semi-circle is always 90°.

Hence ∠ACB = 90°

Now let AC = x and BC = y

Then by applying the Pythagoras theorem to the right-angled triangle ABC, we get

(2r)² = x²+y²

⇒ y² = 4r²-x²

Now we know area of ΔABC is

Now substituting value from equation (i), we get

Now we will find the first derivative of the area, we get

Applying the product rule of differentiation and taking out the constant terms, we get

Applying the power rule of differentiation, we get

Now to find the critical point we will equate the first derivative to 0, i.e.,

A’ = 0

⇒ 2r²-x² = 0

⇒2r² = x²

Or

Now we know, r≠0

Hence and x = r√2

Now we will find the second derivative of the area equation, this can be done by again differentiating equation (iii), we get

Taking out the constant terms and applying the product rule of differentiation, we get

Applying the power rule of differentiation, we get

For in above equation, we get

Hence for , the area of ΔABC is maximum

Now we will find the maximum value by substituting in equation (i), we get

⇒ y = r√2 = x

i.e., the two sides of the ΔABC are equal

Hence the area of Δ ABC is maximum, when it is isosceles

Hence proved