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

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}

Now,

\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}

 

Posted by

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