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  • Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise Multiple Choice question, question 19.

Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise Multiple Choice question, question 19.

Answers (1)

Answer: option(d)  \frac{2}{3}

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

Given: h=R+\sqrt{R^2-r^2}

Solution:

We have,

Solution:

AB=2r

OC=r

CD =R+h                                                     

We have,

h=R+\sqrt{R^2-r^2}

h-R=\sqrt{R^2-r^2}    

Squaring on both sides,

(h-R)^=(\sqrt{R^2-r^2})^2

h^2+R^2-2hR=R^2-r^2

r^2=2hR-h^2                                                                                                 …(i)

Now,

V=\frac{1}{3}\pi r ^2h

V=\frac{\pi}{3}(2hR-h^2)h                        …using equation(i)

\begin{aligned} &V=\frac{\pi}{3}\left(2 h^{2} R-h^{3}\right) \\ &\frac{d V}{d h}=\frac{\pi}{3}\left(4 h R-3 h^{2}\right) \end{aligned}                                                       

For maxima and minima  \frac{dV}{dh}=0

\begin{aligned} &\Rightarrow \frac{\pi}{3}\left(4 h R-3 h^{2}\right)=0 \\ &\Rightarrow 4 h R-3 h^{2}=0 \\ &\Rightarrow 3 h^{2}=4 h R \\ &\Rightarrow h=\frac{4 R}{3} \end{aligned}

Now,

\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h)=\frac{\pi}{3}\left(4 R-6 \times \frac{4 R}{3}\right)=\frac{\pi}{3} 4 R\left(1-\frac{6}{3}\right)=\frac{\pi}{3} 4 R\left(\frac{-3}{3}\right)=-\frac{4 \pi R}{3}<0

Volume is maximum, when  h=\frac{4R}{3}

h=\frac{2(2R)}{3}

\frac{h}{2R}=\frac{2}{3}

Posted by

infoexpert24

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