Question

Of all the closed cylindrical cans (right circular), which enclosed a given volume 100 cubic centimeters, find the dimensions of the minimum surface area.

Answer 1

Let r cm be the radius of base and h cm be the height of the cylindrical can. Let its volume be V and S its total surface area. Then, $V=100c{m}^3$
$\Rightarrow \pi r^{2}h=100$ (given)
$\Rightarrow h=\frac{100}{\pi r^2}\dots \left(i\right)$
Also, $S=2\pi r^2+2\pi rh\dots \left(ii\right)$
In Eq. (ii), there are two independent variables r and h. We eliminate h, by putting the value of h in Eq. (ii) from Eq. (i), we get
$S=2\pi r^2+2\pi r\left(\frac{100}{\pi r^2}\right)=2\pi r^2+\frac{200}{r}\dots \left(iii\right)$
On differentiating Eq. (iii) w.r.t.r, we get $\frac{dS}{dr}=4\pi r-\frac{200}{r^2}\dots \left(iv\right)$
Now, for maxima or minima put $\frac{dS}{dr}=0\Rightarrow 4\pi r=\frac{200}{r^2}$
$\Rightarrow r^3=\frac{200}{4\pi }\Rightarrow r={\left(\frac{50}{\pi }\right)}^{\frac{1}{3}}$

On differentiating Eq. (iv) w.r.t. r, we get $\frac{d^{2}S}{d{r}^2}=4\pi +\frac{400}{r^3}$
AT $r={\left(\frac{50}{\pi }\right)}^{\frac{1}{3}},\frac{d^{2}S}{d{r}^2}=\frac{400}{\left(\frac{50}{\pi }\right)}+4\pi =\frac{400}{50}\times \pi +4\pi =8\pi +4\pi =12\pi >0$
$\therefore$ By second derivative test. the surface area is minimum when the radius of the cylinder is ${\left(\frac{50}{\pi }\right)}^{\frac{1}{3}}$.

Hence, S is minimum when $r={\left(\frac{50}{\pi }\right)}^{\frac{1}{3}}$
On putting value of r is Eq. (i), we get
$h=\frac{100}{\pi {\left(\frac{50}{\pi }\right)}^{\frac{2}{3}}}=2\left(\frac{50}{\pi }\right){\left(\frac{50}{\pi }\right)}^{\frac{-2}{3}}=2{\left(\frac{50}{\pi }\right)}^{\frac{1}{3}}cm$