A closed rectangular box is made of aluminium sheet of negligible thickness and the length of the box is twice its breadth. If the capacity of the box is $243\, cm^3$, compute the dimensions of the box of least surface area.

9 cm × 4.5 cm × 6 cm

The correct answer is Option (2) → 9 cm × 4.5 cm × 6 cm

Let x cm be the breadth of the box, then its length is 2x cm. Let h cm be its height, then

$2x.x.h = 243$ (given) $⇒ h =\frac{243}{2x^2}$   ...(i)

Let S be the surface area of the box, then

$S = 2(2x.x + 2x.h + x.h) = 2(2x^2 + 3xh)$

$=2(2x^2+3x.\frac{243}{2x^2})$  (using (i))

$=4x^2+\frac{729}{x}$.

Diff. it w.r.t. x, we get

$\frac{dS}{dx}=8x-\frac{729}{x^2}$ and $\frac{d^2S}{dx^2}=8-729.\frac{(-2)}{x^2}=8+\frac{1458}{x^3}$.

Now $\frac{dS}{dx}=0⇒8x-\frac{729}{x^2}=0⇒8x^3-729=0⇒x^3=\frac{729}{8}⇒x=\frac{9}{2}$

At $x=\frac{9}{2},\frac{d^2S}{dx^2}=8+1458×(\frac{2}{9})^3=8+16=24>0$

⇒ S is minimum at $x=\frac{9}{2}$.

For $x=\frac{9}{2},h=\frac{243}{2}×(\frac{2}{9})^2=\frac{243}{2}.\frac{4}{81}=6\,cm$

Therefore, the dimensions of the box are $\frac{9}{2}$ cm, 9 cm and 6 cm.