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Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 11.

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Hint: For maxima or minima we must have f'(x)=0

Given: From the question, the area of triangle,

A= \frac{1}{2}ab\sin \theta

Solution: A(\theta)= \frac{1}{2}ab\sin \theta

A'(\theta)= \frac{1}{2}ab\cos \theta

For maxima and minima A'(\theta)= 0

\frac{1}{2}ab \cos \theta =0

\cos \theta =0

\theta =\frac{\pi}{2}

Also A''(\theta)=\frac{1}{2}ab \sin \theta or A''\left ( \frac{\pi}{2} \right )=-\frac{1}{2}ab \sin \frac{\pi}{2}

A^{\prime \prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2} a b<0


\theta =\frac{\pi}{2} is teh point of maxima

The maximum area of triangle \frac{1}{2}ab \sin \frac{\pi}{2}

= \frac{ab}{2}


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