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  • Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 27 maths.

Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 27 maths.

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500 \pi cm^3 

Hint: For the maximum value of v, we must have \frac{dv}{dh}=0

Given: Let height, radius of base and volume of cylinder be h,r and v respectively

Solution: \frac{h^2}{4}+r^2=R^2

h^2=4(R^2-r^2)

r^2=R^2-\frac{h^4}{4}               .......(1)

Now, v=\pi r^2h

v=\pi \left [ hR^2-\frac{h^3}{4} \right ]                      ......... from (1)

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

\begin{aligned} &\frac{d^{2} v}{d h^{2}}=\frac{-3 \pi h}{2}=\frac{-3 \pi}{2} \times \frac{2 R}{\sqrt{3}} \\ &\frac{d^{2} v}{d h^{2}}=\frac{-3 \pi R}{\sqrt{3}}<0 \end{aligned}

Volume of maximum h =\frac{2R}{\sqrt{3}}

Maximum volume =\pi h =\left ( R^2-\frac{h^2}{4} \right )

\begin{aligned} &=\pi \times \frac{2 R}{\sqrt{3}}\left(R^{2}-\frac{4 R^{2}}{12}\right) \\ &=\frac{2 \pi R}{\sqrt{3}}\left(\frac{8 R^{2}}{12}\right)=\frac{4 \pi R^{3}}{3 \sqrt{3}}=\frac{4 \pi(5 \sqrt{3})^{3}}{3 \sqrt{3}} \\ &=500 \pi \mathrm{cm}^{3} \end{aligned}

 

 

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