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

Toolbox:
• To obtain the absolute maxima or minima for the function $f(x)$
• (i) Find $f'(x)$ and put $f'(x)=0$
• (ii) Obtain the points from $f'(x)=0$
• (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$\Delta ABC$ is a right angled
triangle, right angled at $C$
$( \because$ angle in a semicircle is $90^{\circ})$
Hence the area of the $\Delta ABC$ is
$A = \large\frac{1}{2} \times AC \times BC$
Let $D$ be the diameter and $BC = x$
$\therefore AC=\sqrt{D^2-x^2}$
$\therefore A= \large\frac{1}{2}=\sqrt{D^2-x^2} \times x$
Step 2
Differentiating w.r.t $x$ we get,
$\large\frac{dA}{dx}=\large\frac{1}{2} \bigg[ \sqrt{D^2-x^2}.1 + x \large\frac{1}{2\sqrt{D^2-x^2}} (-2x) \bigg]$
$=\large\frac{1}{2} \bigg[ \large\frac{D^2-x^2-x^2}{\sqrt{D^2-x^2}} \bigg] = \large\frac{1}{2} \bigg[ \large\frac{D^2-2x^2}{\sqrt{D^2-x^2}} \bigg]$
To find the maximum or minimum area
$\large\frac{dA}{dx}=0$
i.e., $\large\frac{1}{2} \bigg[ \large\frac{D^2-2x^2}{\sqrt{D^2-x^2}} \bigg]=0$
$\Rightarrow D^2=2x^2$
$D = x\sqrt 2$
Step 3
To prove that the triangle has maximum area let us differentiate $\large\frac{dA}{dx}$ again
$\large\frac{d^2A}{dx^2}=\large\frac{\sqrt{D^2-x^2}(-4x)-(D^2-2x^2) \large\frac{1}{2\sqrt{D^2-x^2}}(-2x)}{(D^2-x^2)}$
$\large\frac{d^2A}{dx^2} = \large\frac{\sqrt{D^2-x^2}[-3x]}{D^2-x^2} <0$
hence this proves that the triangle has maximum value.
$AC^2=D^2-x^2$
$=2x^2-x^2$
$AC^2=x^2$
$\Rightarrow AC=x$
$AC=BC=x$
Hence the triangle has maximum value when it is an isosceles triangle.