What is the minumum $cm^2$ tin which you need to create $1L$ closed cylinder
Any hint?
it is $\pi r^2h=1$
$2\pi r^2+2\pi rh =2(\frac {r+h}{hr})$
edit:
$h=\frac{1}{\pi r^2}$
$f(r)=2\pi r^2+2\pi r\frac{1}{\pi r^2}=2\pi r^2+\frac{2}{r}$
$f'(r)=0 \rightarrow 4\pi r=\frac{2}{r^2}$
$r= {\dfrac {1} {\sqrt[3]{2\pi} }}$
$f(r)=2\pi{\dfrac {1} {\sqrt[3]{4\pi^2} }}+ 2\sqrt[3]{2\pi} $
$f(r)=\sqrt[3]{2\pi}+ 2\sqrt[3]{2\pi}= 3\sqrt[3]{2\pi}$
This answer isnt in the options. Can u tell me where is the mistake?
As you have Calculated $\displaystyle f(r) = 2\pi r^2+\frac{2}{r}\;,$ Here $\bf{r(radius)>0}$
So Using $\bf{A.M\geq G.M\;,}$ We get $\displaystyle \frac{\left[2\pi r^2+\frac{1}{r}+\frac{1}{r}\right]}{3}\geq \left(2\pi r^2\cdot \frac{1}{r}\cdot \frac{1}{r}\right)^{\frac{1}{3}}$
So we get $\displaystyle2\pi r^2+ \frac{1}{r}+\frac{1}{r}\geq 3\cdot \left(2\pi \right)^{\frac{1}{3}}$
And equality hold when $\displaystyle 2\pi r^2 = \frac{1}{r}\Rightarrow r= \left(\frac{1}{2\pi}\right)^{\frac{1}{3}}$
So Minimum Surface area $\displaystyle f(r)$ occur at $\displaystyle r= \left(\frac{1}{2\pi}\right)^{\frac{1}{3}}$
So we get $\displaystyle S \left[\left(\frac{1}{2\pi}\right)^{\frac{1}{3}}\right] = 2\pi \cdot \left(\frac{1}{2\pi}\right)^{\frac{2}{3}}+\frac{2}{ \left(\frac{1}{2\pi}\right)^{\frac{1}{3}}} = \left(2\pi\right)^{\frac{1}{3}}+2\cdot \left(2\pi\right)^{\frac{1}{3}} = 3\cdot \left(2\pi\right)^{\frac{1}{3}}$
May be Option Given is Wrong.