Question

An open box with a square base is to be made out of a given quantity of cardboard of area $c^2$ square units. Show that the maximum volume of the box is $\frac{c^3}{6\sqrt{3}}$ cubic units.

Answer 1

The area of the square piece is given to be $c^2$ square units.

Since the base is a square, let the length and breadth of the resulting box be $l$ and the height be $h$.

Therefore, $l^2+4lh=c^2$, equating the areas.

Also, the volume of the box is thus given by length $\times$ breadth $\times$ height = $l^{2}h$

We can write the volume only in terms of $l$ as $l^2\times \frac{c^2-l^2}{4l}=\frac{l{c}^2-l^3}{4}$

Differentiating this w.r.t $l$ and equating it to zero, we get $c^2-3l^2=0$

$\Rightarrow l=\sqrt{\frac{c^2}{3}}$

The volume thus becomes $\frac{l{c}^2-l^3}{4}=\sqrt{\frac{c^2}{3}}\times \frac{c^2-\frac{c^2}{3}}{4}$

$=\frac{c}{\sqrt{3}}\times \frac{c^2}{6}$

$=\frac{c^3}{6\sqrt{3}}$