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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.

Answers (1)

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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