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

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

Answers (1)

\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}

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