A covered box of volume 72 cm3 and the base sides in a ratio of 1:2 is to be made. The length of all sides so that the total surface area is the least possible is

Let $l, b, h$ be the dimensions $l = 2 b$ , so $V = l b h = 2 b^{2} h$

$\Rightarrow 72 = 2 b^{2} h \Rightarrow h = \frac{36}{b^{2}}$

The surface area $S = 2 (l b + b h + l h)$

$= 2 (2 b^{2} + \frac{3 b}{b} + 2 b \times \frac{36}{b^{2}})$

$= 2 (2 b^{2} + \frac{108}{b})$

$= 4 (b^{2} + \frac{54}{b})$

$\frac{d S}{d b} = 4 (2 b - \frac{54}{b^{2}}), \frac{d S}{d b}$ is zero is $b = 3$ and

$\frac{d^{2} S}{d b^{2}} = 4 (2 + \frac{108}{b^{3}}) > 0$

Hence $S$ is minimum when $b$ = 3. So the dimensions are

$6, 3, \frac{36}{9} = 6, 3, 4$

A covered box of volume 72 ${cm}^{3}$ and the base sides in a ratio of 1:2 is to be made. The length of all sides so that the total surface area is the least possible is