UPSC NDA 2 2017 Mathematics Solved Paper

Concept:For a cylinder without a lid, the volume is maximized when its height equals its radius, given a fixed surface area.Explanation:Let the radius be $r$ and height be $h$ of the cylindrical jar.The total surface area (excluding the lid) is $A = \pi r^2 + 2\pi r h$.The volume is $V = \pi r^2 h$.From the surface area equation, solve for $h$: $h = \frac{A}{2\pi r} - \frac{r}{2}$.Substitute this into the volume: $V = \pi r^2 \left( \frac{A}{2\pi r} - \frac{r}{2} \right) = \frac{Ar}{2} - \frac{\pi r^3}{2}$.To maximize $V$, differentiate with respect to $r$ and set derivative to zero: $\frac{dV}{dr} = \frac{A}{2} - \frac{3\pi r^2}{2} = 0 \Rightarrow A = 3\pi r^2$.Now substitute $A = 3\pi r^2$ back into the expression for $h$: $h = \frac{3\pi r^2}{2\pi r} - \frac{r}{2} = \frac{3r}{2} - \frac{r}{2} = r$.Thus, at maximum volume, $h = r$. The diameter $d = 2r$, so $d = 2h$.Given that the diameter $d$ is $k$ times the height $h$, we have $d = k h \Rightarrow 2h = k h \Rightarrow k = 2$.Answer:$k = 2$, which corresponds to option B.