Question

A tin must be shaped as a cylinder with I $d{m}^3$ capacity. What must be the radius of its bases for the area of the tin sheet used to make the tin to be at a minimum?

Answer 1

Capacity : $1d{m}^3$, let radius $=r$

height $=h$

volume $=\pi r^{2}h=1d{m}^3$

$h=\frac{1}{\pi r^2}$

Total Area $=2\pi r^2+2\pi rh=2\pi r(r+\frac{1}{\pi r^2})=2\pi r^2+\frac{2}{r}$

For minimum area, $\frac{d}{dr}\left(Area\right)=0=4\pi r-\frac{2}{r^2}\Rightarrow r^3=\frac{1}{2\pi }$

$\therefore r={\left(\frac{1}{2\pi }\right)}^{\frac{1}{3}}dm$