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

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

Answers (1)

\frac{ab}{2}

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

i.e,

\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}

 

Posted by

infoexpert24

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