ABC26GN4904 · Volume and Surface Area
Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
The height of a closed cylinder of given volume and the minimum surface area is
(a)equal to its diameter
(b)half of its diameter
(c)double of its diameter
(d)None of these
Answer
Explanation
$V=\pi r^{2} h$ and $S=2 \pi r h+2 \pi r^{2} \Rightarrow S=2 \pi r(h+r)$, where $h=\frac{\mathrm{V}}{\pi r^{2}} \Rightarrow S=2 \pi r\left(\frac{\mathrm{V}}{\pi r^{2}}+r\right)=\frac{2 \mathrm{V}}{r}+2 \pi r^{2}$ $$\begin{aligned} & \Rightarrow \frac{d \mathrm{S}}{d r}=\frac{-2 \mathrm{V}}{r^{2}}+4 \pi r \text { and } \frac{d^{2} \mathrm{S}}{d r^{2}}=\left(\frac{4 \mathrm{V}}{r^{3}}+4 \pi\right)>0 \\ & \therefore S \text { is minimum when } \frac{d \mathrm{S}}{d r}=0 \Leftrightarrow \frac{-2 \mathrm{V}}{r^{2}}+4 \pi r=0 \Leftrightarrow V \\ & =2 \pi r^{3} \Leftrightarrow \pi r^{2} h=2 \pi r^{3} \Leftrightarrow h=2 r . \end{aligned}$$
Open in whiteboard · Browse this chapter in the app