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Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 19.

Answers (1)

h = \sqrt{2}r

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

Given: Surface area of conical tank S=\pi r \sqrt{r^2+h^2}

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

Solution: V= \frac{1}{3}\pi r^2 h

\begin{aligned} &h=\frac{3 V}{\pi r^{2}} \\ &\therefore S=\pi r \sqrt{r^{2}+\left(\frac{3 V}{\pi r^{2}}\right)^{2}} \\ &=\frac{1}{r} \sqrt{\pi^{2} r^{6}+9 V^{6}} \end{aligned}


\begin{aligned} &\frac{d S}{d r}=\frac{d}{d r}\left[\frac{1}{r} \sqrt{\pi^{2} r^{6}+9 V^{6}}\right] \\ &=\frac{1}{r} \frac{6 \pi^{2} r^{5}}{\sqrt{\pi^{2} r^{6}+9 V^{6}}}-\frac{\sqrt{\pi^{2} r^{6}+9 V^{6}}}{r^{2}} \end{aligned}

For minima \frac{dS}{dr}=0

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

Substitute V value in equation (1)

\begin{aligned} &2 \pi^{2} r^{6}=9\left(\frac{1}{3} \pi r^{2} h\right)^{6} \\ &2 \pi^{2} r^{6}=\pi^{2} r^{4} h^{2} \\ &2 r^{2}=h^{2} \\ &h=\sqrt{2} r \end{aligned}


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