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  • Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 17.

Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 17.

Answers (1)

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

Hint: For maxima and minima value of A, we must have \frac{dA}{dx}=0

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

Solution:  Let the height and radius of the base of the cylinder be h and r

\frac{h^2}{4}+r^2=R^2

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

Volume of the cylinder V =\pi r^2 h

Square on both the sides

V^2=\pi ^2 r^4h^2

V^2=4\pi ^2 r^4 (R^2-r^2)                       ..........from (1)

Now,

\begin{aligned} &Z=4 \pi^{2}\left(r^{4} R^{2}-r^{6}\right) \\ &\frac{d Z}{d x}=4 \pi^{2}\left(4 r^{3} R^{2}-6 r^{5}\right) \\ &\frac{d Z}{d x}=0 \\ &4 \pi^{2}\left(4 r^{3} R^{2}-6 r^{5}\right)=0 \end{aligned}

\begin{aligned} &4 r^{3} R^{2}=6 r^{5} \\ &6 r^{2}=4 R^{2} \\ &r^{2}=\frac{4 R^{2}}{6} \\ &r=\frac{2 R}{\sqrt{6}} \end{aligned}

Subsitute r value in eqn (1)

\begin{aligned} &h=2 \sqrt{R^{2}-\left(\frac{2 R}{\sqrt{6}}\right)^{2}} \\ &=2 \sqrt{\frac{6 R^{2}-4 R^{2}}{6}}=2 \sqrt{\frac{R^{2}}{3}}=\frac{2 R}{\sqrt{3}} \\ &\frac{d^{2} Z}{d x^{2}}=4 \pi^{2}\left(12 r^{2} R^{2}-30 r^{4}\right) \end{aligned}

\begin{aligned} &=4 \pi^{2}\left(12\left(\frac{2 R}{\sqrt{6}}\right)^{2} R^{2}-30\left(\frac{2 R}{\sqrt{6}}\right)^{4}\right) \\ &=4 \pi^{2}\left(8 R^{4}-\frac{80 R^{4}}{6}\right) \end{aligned}

\begin{aligned} &=4 \pi^{2}\left(\frac{48 R^{4}-80 R^{4}}{6}\right) \\ &=4 \pi^{2}\left(\frac{-16 R^{4}}{3}\right)<0 \end{aligned}

Volume of the cylinder is max when, h=\frac{2R}{\sqrt{3}}

 

Posted by

infoexpert24

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