Question

A cylindrical gas container is closed at the top and open at the bottom; if the iron plate of the top is $5/4$ times as thick as the plate forming the cylindrical sides, find the radio of the radius to the height of the cylinder using minimum material for the same capacity.

• A. $\frac{5}{4}$
• B. $\frac{4}{5}$
• C. $\frac{5}{2}$
• D. $\frac{2}{5}$

Answer 1

The correct option is B $\frac{4}{5}$

$V=\pi r^{2}h,$ if k be the thickness of the sides then that of the top will be $\frac{5}{4}k$ $\therefore S=\left(\pi rh\right)k+\left(\pi r^2\right)\frac{5}{4}k$ $orS=2\pi r.\frac{V}{\pi r^2}k+\frac{5}{4}\pi r^{2}k=k(\frac{2V}{r}+\frac{5}{4}\pi r^2)$ Note:We have taken into consideration the actual inaterial used and not as in last question in which the base wasw made from a square sheet thus wasting the material .$\therefore \frac{ds}{dr}=k(\frac{-2V}{r^2}+\frac{5}{2}\pi r)(5\pi +\frac{5}{2})=+ive.$ when $r^3=\frac{4V}{5\pi }or5\pi r^3=4\pi r^{2}h$ $\therefore \frac{r}{h}=\frac{4}{5}.$