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.

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Solution

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, $l^{2} + 4 l h = c^{2}$ , 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 $l^{2} \times \frac{c^{2} - l^{2}}{4 l} = \frac{l c^{2} - l^{3}}{4}$
Differentiating this w.r.t $l$ and equating it to zero, we get $c^{2} - 3 l^{2} = 0$
$\Rightarrow l = \sqrt{\frac{c^{2}}{3}}$
The volume thus becomes $\frac{l c^{2} - l^{3}}{4} = \sqrt{\frac{c^{2}}{3}} \times \frac{c^{2} - \frac{c^{2}}{3}}{4}$
$= \frac{c}{\sqrt{3}} \times \frac{c^{2}}{6}$
$= \frac{c^{3}}{6 \sqrt{3}}$

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An open box with a square base is to be made out of a given quantity of cardboard of area c2 square units. Show that the maximum volume of the box is c36√3 cubic units.

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.