#### Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 21 maths.

$r =\left ( \frac{3V}{\pi \sqrt{2}} \right )^{\frac{1}{3}}$   or $V=\frac{\pi r^2 \sqrt{2}}{3}$

Hint: For maxima or minima f'(x) must be zero f'(x) = 0

Given: Volume $=\frac{1}{3}\pi r^2 h$

Solution: $V=\frac{1}{3}\pi r^2 h,h =\frac{3V}{\pi r^2}$

Slant height, $l = \sqrt{h^2+r^2}$

\begin{aligned} &=\sqrt{\left(\frac{3 V}{\pi r^{2}}\right)^{2}+r^{2}} \\ &=\sqrt{\left(\frac{9 V^{2}}{\pi^{2} r^{4}}\right)+r^{2}} \\ &I=\sqrt{\frac{9 V^{2}+\pi^{2} r^{6}}{\pi^{2} r^{4}}}=\frac{\sqrt{9 V^{2}+\pi^{2} r^{6}}}{\pi r^{2}} \end{aligned}

Curved surface area, $e = \pi r l$

\begin{aligned} &C(r)=\frac{\pi r \sqrt{9 V^{2}+\pi^{2} r^{6}}}{\pi r^{2}} \\ &=\frac{\sqrt{9 V^{2}+\pi^{2} r^{6}}}{r} \end{aligned}

\begin{aligned} &C^{\prime}(r)=\frac{\frac{r \times 6 \pi^{2} r^{5}}{2 \sqrt{9 V^{2}+\pi^{2} r^{6}}}-\sqrt{9 V^{2}+\pi^{2} r^{6}}}{r^{2}} \\ &=\frac{\left[\frac{3 \pi^{2} r^{6}-\left(9 V^{2}+\pi^{2} r^{6}\right)}{\sqrt{9 V^{2}+\pi^{2} r^{6}}}\right]}{r^{2}} \end{aligned}

\begin{aligned} &=\frac{3 \pi^{2} r^{6}-9 V^{2}-\pi^{2} r^{6}}{r^{2} \sqrt{9 V^{2}+\pi^{2} r^{6}}} \\ &=\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{9 V^{2}+\pi^{2} r^{6}}} \end{aligned}

$C'(r)=0$

\begin{aligned} &\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{9 V^{2}+\pi^{2} r^{6}}}=0 \\ &2 \pi^{2} r^{6}=9 V^{2} \\ &V^{2}=\frac{2 \pi^{2} r^{6}}{9}, V=\sqrt{\frac{2 \pi^{2} r^{6}}{9}} \end{aligned}

$V=\frac{\pi r^{3} \sqrt{2}}{3} \text { or } r=\left(\frac{3 V}{\pi \sqrt{2}}\right)^{\frac{1}{3}}$

So,

$h =\frac{3}{\pi r^2}\times \frac{\pi r^3 \sqrt{2}}{3}$

$h =r \sqrt{2}$

$\frac{h }{r}= \sqrt{2}$

$\cot \theta = \sqrt{2}, \theta = \cot^{-1}\sqrt{2}$

Since, $r <\left ( \frac{3V}{\pi \sqrt{2}} \right )^{\frac{1}{3}}, C'(r)<0$ and for $r >\left ( \frac{3V}{\pi \sqrt{2}} \right )^{\frac{1}{3}}, C'(r)>0$

Curved surface area for  $r =\left ( \frac{3V}{\pi \sqrt{2}} \right )^{\frac{1}{3}}$   or $V=\frac{\pi r^2 \sqrt{2}}{3}$

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