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 diameter and C is any point on the circle with radius r.
∠ACB = 90o [angle in the semi-circle is 90o]
Let AC = x
$B C=\sqrt{\bar{A} B^{2}-A C^{2}}$
$\mathrm{BC}=\sqrt{(2 r)^{2}-x^{2}} \Rightarrow \mathrm{BC}=\sqrt{4 r^{2}-x^{2}}$$\ldots(i)$
Now area of $\triangle \mathrm{ABC}, \mathrm{A}=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BC}$
$\Rightarrow A=\frac{1}{2} x \cdot \sqrt{4 r^{2}-x^{2}}$
Squaring on both the sides, we get
$\mathrm{A}^{2}=\frac{1}{4} x^{2}\left(4 r^{2}-x^{2}\right)$
Let $\mathrm{A}^{2}=\mathrm{Z}$
$\mathrm{SO}$ $Z=\frac{1}{4} x^{2}\left(4 r^{2}-x^{2}\right) \Rightarrow Z=\frac{1}{4}\left(4 x^{2} r^{2}-x^{4}\right)$
Differentiating both sides w.r.t. $x$, we get
$\frac{d Z}{d x}=\frac{1}{4}\left[8 x r^{2}-4 x^{3}\right]$ $\ldots(i i)$
For local maxima and local minima $\frac{d Z}{d x}=0$
$x \neq 0 \quad$ and $\quad 2 r^{2}-x^{2}=0$
$\Rightarrow x^{2}=2 r^{2} \Rightarrow x=\sqrt{2} r=\mathrm{AC}$
Now from eq. (i) we have
$\mathrm{BC}=\sqrt{4 r^{2}-2 r^{2}} \Rightarrow \mathrm{BC}=\sqrt{2 r^{2}} \Rightarrow \mathrm{BC}=\sqrt{2} r$
Thus, $\quad A C=B C$
Hence, $\triangle \mathrm{ABC}$ is an isosceles triangle.
Differentiating eq. (ii) w.r.t. $x$, we get $\frac{d^{2} z}{d x^{2}}=\frac{1}{4}\left[8 r^{2}-12 x^{2}\right]$
Put $x=\sqrt{2} r$
$\therefore \frac{d^{2} Z}{d x^{2}}=\frac{1}{4}\left[8 r^{2}-12 \times 2 r^{2}\right]=\frac{1}{4}\left[8 r^{2}-24 r^{2}\right]$
Therefore, the area of Δ ABC is minimum when it is an isosceles triangle.