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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

$\begin{array}{1 1} \frac{10}{\pi}\big(\frac{\pi}{50}\big)^{\frac{2}{3}}cm \\ \frac{100}{\pi}\big(\frac{\pi}{20}\big)^{\frac{2}{3}}cm \\ \frac{100}{\pi}\big(\frac{\pi}{50}\big)^{\frac{2}{3}}cm \\ \frac{100}{\pi}\big(\frac{-\pi}{50}\big)^{\frac{2}{3}}cm \end{array} $

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  • Volume =$\pi r^2h$
Step 1:
Let $r$ be the radius and $h$ be the height of cylindrical can.
Volume=$\pi r^2h$
$h=\large\frac{100}{\pi r^2}$
Total surface area of the can $S=2\pi r h+2\pi r^2$
$\Rightarrow 2\pi r\big(\large\frac{100}{\pi r^2}\big)$$+2\pi r^2$
$\Rightarrow \large\frac{200}{r}$$+2\pi r^2$
Step 2:
$\large\frac{dS}{dr}=-\large\frac{200}{r^2}$$+4\pi r$
$\Rightarrow \large\frac{-200+4\pi r^3}{r^2}$
Now $\large\frac{dS}{dr}$$=0$
$\Rightarrow \large\frac{-200+4\pi r^2}{r^2}$$=0$
$4\pi r^3-200=0$
$\pi r^3=50$
Step 3:
Also $\large\frac{d^2S}{dx^2}=\frac{400}{r^3}$$+4\pi$
At $r=\big[\large\frac{50}{\pi}\big]^{\Large\frac{1}{3}}$
$\Rightarrow S$ is minimum or least when $r=\big(\large\frac{50}{\pi}\big)^{\Large\frac{1}{3}}$
Hence the total surface area is least when radius of base is $\big[\large\frac{50}{\pi}\big]^{\Large\frac{1}{3}}$cm.
$h=\large\frac{100}{\pi r^2}$
$\Rightarrow \large\frac{100}{\pi}\big(\large\frac{\pi}{50}\big)^{\Large\frac{2}{3}}$cm.
answered Aug 8, 2013 by sreemathi.v

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