$A B$ is the diameter of a circle and $C$ is any point on the circle. Show that the area of triangle $A B C$ is maximum, when it is an isosceles triangle.

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Solution

Since $A B$ is a diameter of the circle.
$∴ A B = 2 R$
Let $A C = x B C = y$
$∴ In right angled △ A C B, A C^{2} + B C^{2} = A B^{2} \Rightarrow x^{2} + y^{2} = {(2 R)}^{2} \Rightarrow x^{2} + y^{2} = 4 R^{2} \Rightarrow y^{2} = 4 R^{2} - x^{2} . . . (i)$
Now, area of the $△ A C B = \frac{1}{2} \times A C \times B C$
$\Rightarrow A = \frac{1}{2} \times x \times y \Rightarrow A^{2} = \frac{1}{4} \times x^{2} \times y^{2}$
$Suppose S = A^{2} \Rightarrow A^{2} = \frac{1}{4} \times x^{2} \times (4 R^{2} - x^{2}) [Using (i)] S = x^{2} R^{2} - \frac{1}{4} x^{4},$ $\frac{d S}{d x} = 2 R^{2} x - x^{3} . . . (i i)$
Now, $\frac{d S}{d x} = 0$
$\Rightarrow 2 R^{2} x - x^{3} = 0 \Rightarrow (2 R^{2} - x^{2}) = 0 \Rightarrow x^{2} = 2 R^{2} [∵ x \neq 0] \Rightarrow x = \sqrt{2} R (i i i)$

Again, differentiating both sides of $(i i)$ w.r.t. $x$ , we have

$\frac{d^{2} S}{d x^{2}} = (2 R^{2} - 3 x^{2}) = 2 R^{2} - 3 (2 R^{2}) [U s i n g (i i i)] = 2 R^{2} - 6 R^{2} = - 4 R^{2} < 0$
$∴$ For $x = \sqrt{2} R$ ,
Area of the triangle is maximum Also, for $x = \sqrt{2} R,$
From $(i)$ ,
$y$ = $\sqrt{4 R^{2} - 2 R^{2}} = \sqrt{2 R^{2}} = \sqrt{2} R i . e ., x = y = \sqrt{2} R$
Hence, $△ A B C$ is an isosceles triangle.

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