Question

Show that a cylinder of a given volume, which is open at the top, has minimum total surface area when its height is equal to the radius of its base.

Answer 1

Let r be the radius and h be the height of a cylinder of given volume V. Then,
V = \[\pi r^2 h\]

⇒ \[h = \frac{V}{\pi r^2}\]     ...(i)

Let S be the total surface area of the cylinder. Then,

\[S = 2\pi r h + \pi r^2\]

⇒ \[S = 2\pi r\left( \frac{V}{\pi r^2} \right) + \pi r^2\]        {Using (i)}

⇒ \[S = \frac{2V}{r} + \pi r^2\]

⇒ \[\frac{dS}{dr} = - \frac{2V}{r^2} + 2\pi r\]       ...(ii)

For maximum or minimum,

\[\frac{dS}{dr} = 0\]

⇒ \[- \frac{2V}{r^2} + 2\pi r = 0\]

⇒ \[\frac{2V}{r^2} = 2\pi r\]

⇒ \[V = \pi r^3\]

⇒ \[\pi r^2 h = \pi r^3 \Rightarrow h = r\]

Differentiating (ii) w.r.t r, we get:

\[\frac{d^2 S}{d r^2} = \frac{6V}{r^3} + 2\pi > 0\] 

Hence, S is minimum when h = r, i.e. when the height of the cylinder is equal to the radius of the base.