Show that a cylinder of given volume, open at the top, has minimum total surface area if its height is - Maths - Application of Derivatives

Question

Show that a cylinder of given volume, open at the top, has
minimum total surface area if its height is equal to radius of the
base (2009c)

Santosh Kumar · Accepted Answer

Let r be the radius and h be the height of the cylinder of given volume v.

Now,

$v = π r^{2} h \Rightarrow h = \frac{v}{π r^{2}} ........ (i)$

As the cylinder is open at the top.

So, total surface area, s = 2πrh + πr²

$\Rightarrow s = 2 π r (\frac{v}{π r^{2}}) + π r^{2} [using equation (i)] \Rightarrow s = \frac{2 v}{r} + π r^{2} \Rightarrow \frac{d s}{d r} = \frac{- 2 v}{r^{2}} + 2 π r ....... (ii)$

For total surface area to be minimum,

$\frac{d s}{d r} = 0 \Rightarrow \frac{- 2 v}{r^{2}} + 2 π r = 0 \Rightarrow v = π r^{3} \Rightarrow π r^{2 h} = π r^{3} \Rightarrow h = r$

Differentiating equation (ii) with respect to r,

$\frac{d^{2} s}{d r^{2}} = \frac{4 v}{r^{3}} + 2 π \Rightarrow {(\frac{d^{2} s}{d r^{2}})}_{r = h} = \frac{4 v}{h^{3}} + 2 π > 0$

Thus, surface area is minimum, when h = r, i.e., when the height of the cylinder is equal to the radius of the base.

Show that a cylinder of given volume, open at the top, has minimum total surface area if its height is equal to radius of the base. (2009c)