#### Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 22 maths.

$\theta = \frac{\pi}{6}$

Hint: For maxima or minima f'(x) must be zero f'(x) = 0

Given:

Solution:

Let ABC be an isoceles triangle inscribed in the circle with the radius a such that

AB=AC

$AD =AO+OD =a + a\cos \theta = a (1+2\cos \theta)$

$BC =2BD =2a \sin 2\theta$

Area of triangle ACA $=\frac{1}{2}BC\times AD$

\begin{aligned} &A(\theta)=\frac{1}{2} \times 2 a \sin 2 \theta \times a(1+\cos 2 \theta) \\ &=a^{2} \sin 2 \theta(1+\cos 2 \theta) \\ &A(\theta)=a^{2} \sin 2 \theta+a^{2} \sin 2 \theta \cos 2 \theta \\ &A(\theta)=a^{2} \sin 2 \theta+\frac{a^{2} \sin 4 \theta}{2} \end{aligned}

\begin{aligned} &A^{\prime}(\theta)=2 a^{2} \cos 2 \theta+\frac{4 a^{2} \cos 4 \theta}{2} \\ &A^{\prime}(\theta)=2 a^{2} \cos 2 \theta+2 a^{2} \cos 4 \theta \\ &=2 a^{2}(\cos 2 \theta+\cos 4 \theta) \end{aligned}

\begin{aligned} &A^{\prime}(\theta)=0 \\ &2 a^{2}(\cos 2 \theta+\cos 4 \theta)=0 \\ &\cos 2 \theta+\cos 4 \theta=0 \\ &\cos 2 \theta=-\cos 4 \theta \\ &=\cos (\pi-4 \theta) \\ &2 \theta=\pi-4 \theta \\ &6 \theta=\pi \\ &\theta=\frac{\pi}{6} \\ &A^{\prime \prime}(\theta)=2 a^{2}(-\sin 2 \theta-\sin 4 \theta) \\ &=-2 a^{2}(\sin 2 \theta+\sin 4 \theta)<0 \\ &at \ \ \theta=\frac{\pi}{6} \end{aligned}