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, the radio of the radius to the height of the cycliner using minimum material for the same capacity is

Let x be the radius and y the height of the cylindrical gas container. Also, let k be the thickness of the plates forming the cylindrical sides. Therefore,  the thickness of the plate forming the top will be $5k/4$ . Capacity of the vessel = volume of cylinder 

$=\pi x^{2}y=V\left(\text{Given}\right)\therefore y=V/\left(\pi x^2\right)$  (1)

Now, the volume $V_1$ of the iron plate used for construction of the container is given by the container is given by $V_1=\pi (x+k{)}^2(y+5k/4)-\pi x^{2}y$

$\frac{d{V}_1}{dx}=0 \Rightarrow x=[4V/(5\pi ){]}^{1/3}$

For this value of x, $d^2{V}_1/d{x}^2 \text{is +ve}$

Hence, $V_1$ is minimum when $x=[\frac{4V}{5\pi }{]}^{1/3}$.

Now,    $x=[\frac{4V}{5\pi }{]}^{1/3}$  $\Rightarrow$  $5\pi x^3=4V=4\pi x^{2}y \Rightarrow x/y=4/5$.

Hence the required ratio is  $4: 5$