If $AB$ is a diameter of a circle and $C$ is any point on the circle, then show that the area of $\Delta ABC$ is maximum, when it is isosceles.

We have, $AB = 2r$

and $\angle ACB = {90^\circ }\quad$ [since, angle in the semi-circle is always ${90^\circ }$]

Let $AC = x$ and $BC = y$

$\therefore {(2r)^2} = {x^2} + {y^2}$

$\Rightarrow$ ${y^2} = 4{r^2} - {x^2}$

$\Rightarrow$ $y = \sqrt {4{r^2} - {x^2}}$ …….(i)

Now, area of $\Delta ABC,A = \frac{1}{2} \times x \times y$
$= \frac{1}{2} \times x \times {\left( {4{r^2} - {x^2}} \right)^{1/2}}\quad$ [using Eq. (i)]

Now, differentiating both sides w.r.t. $x$, we get
$\frac{{dA}}{{dx}} = \frac{1}{2}\left[ {x \cdot \frac{1}{2}{{\left( {4{r^2} - {x^2}} \right)}^{ - 1/2}} \cdot (0 - 2x) + {{\left( {4{r^2} - {x^2}} \right)}^{1/2}} \cdot 1} \right]$

$= \frac{1}{2}\left[ {\frac{{ - 2{x^2}}}{{2\sqrt {4{r^2} - {x^2}} }} + {{\left( {4{r^2} - {x^2}} \right)}^{1/2}}} \right]$
$= \frac{1}{2}\left[ {\frac{{ - {x^2}}}{{\sqrt {4{r^2} - {x^2}} }} + \sqrt {4{r^2} - {x^2}} } \right]$

$= \frac{1}{2}\left[ {\frac{{ - {x^2} + 4{r^2} - {x^2}}}{{\sqrt {4{r^2} - {x^2}} }}} \right] = \frac{1}{2}\left[ {\frac{{ - 2{x^2} + 4{r^2}}}{{\sqrt {4{r^2} - {x^2}} }}} \right]$

$\Rightarrow$ $\frac{{dA}}{{dx}} = \left[ {\frac{{\left( { - {x^2} + 2{r^2}} \right)}}{{\sqrt {4{r^2} - {x^2}} }}} \right]$

Now, $\frac{{dA}}{{dx}} = 0$

$\Rightarrow$ $- {x^2} + 2{r^2} = 0$
$\Rightarrow$ ${r^2} = \frac{1}{2}{x^2}$

$\Rightarrow$ $r = \frac{1}{{\sqrt 2 }}x$

$\therefore x = r\sqrt 2$

Again, differentiating both sides w.r.t. $x$, we get
$\frac{{{d^2}A}}{{d{x^2}}} = \frac{{\sqrt {4{r^2} - {x^2}} \cdot ( - 2x) + \left( {2{r^2} - {x^2}} \right) \cdot \frac{1}{2}{{\left( {4{r^2} - {x^2}} \right)}^{ - 1/2}}( - 2x)}}{{{{\left( {\sqrt {4{r^2} - {x^2}} } \right)}^2}}}$

$= \frac{{ - 2x\left[ {\sqrt {4{r^2} - {x^2}} + \left( {2{r^2} - {x^2}} \right) \cdot \frac{1}{{2\sqrt {4{r^2} - {x^2}} }}} \right]}}{{{{\left( {\sqrt {4{r^2} - {x^2}} } \right)}^2}}}$

$= \frac{{ - 4x \cdot {{\left( {\sqrt {4{r^2} - {x^2}} } \right)}^2} + \left( {2{r^2} - {x^2}} \right)( - 2x)}}{{2 \cdot {{\left( {4{r^2} - {x^2}} \right)}^{3/2}}}}$

$= \frac{{ - 4x\left( {4{r^2} - {x^2}} \right) + \left( {2{r^2} - {x^2}} \right) \cdot ( - 2x)}}{{2 \cdot {{\left( {4{r^2} - {x^2}} \right)}^{3/2}}}}$

$= \frac{{ - 16x{r^2} + 4{x^3} + \left( {2{r^2} - {x^2}} \right)( - 2x)}}{{2 \cdot {{\left( {4{r^2} - {x^2}} \right)}^{3/2}}}}$

$= \frac{{ - 16\sqrt 2 \cdot {r^3} + 8\sqrt 2 {r^3}}}{{2{{\left( {2{r^2}} \right)}^{3/2}}}} = \frac{{8\sqrt 2 {r^2}[r - 2r]}}{{4{r^3}}}$

$= \frac{{ - 8\sqrt 2 {r^3}}}{{4{r^3}}} = - 2\sqrt 2 < 0$

For $x = r\sqrt 2 ,$ the area of triangle is maximum.

For $x = r\sqrt 2$, $y = \sqrt {4{r^2} - {{(r\sqrt 2 )}^2}} = \sqrt {2{r^2}} = r\sqrt 2$

Since, $x = r\sqrt 2 = y$

Hence it is maximum when the triangle is isosceles.