If an open box with square base is to be made of a given quantity of card board of area ${c^2}$, then show that the maximum volume of the box is $\frac{{{c^3}}}{{6\sqrt 3 }}{\rm{cu}}$ units.

Let us assume that the length of side of the square base of open box be $x$ units and its height be $y$ units.

Therefore area of the metal used $= {x^2} + 4xy$
$\Rightarrow$ ${x^2} + 4xy = {c^2}$ [given]
$\Rightarrow$ $y = \frac{{{c^2} - {x^2}}}{{4x}}$

Now, volume of the box $(V) = {x^2}y$
$\Rightarrow$ $V = {x^2} \cdot \left( {\frac{{{c^2} - {x^2}}}{{4x}}} \right)$
$= \frac{1}{4}x\left( {{c^2} - {x^2}} \right)$

$= \frac{1}{4}\left( {{c^2}x - {x^3}} \right)$

On differentiating both sides w.r.t. $x$, we get
$\frac{{dV}}{{dx}} = \frac{1}{4}\left( {{c^2} - 3{x^2}} \right)$ …….(ii)

Now, $\frac{{dV}}{{dx}} = 0 \Rightarrow {c^2} = 3{x^2}$
$\Rightarrow$ ${x^2} = \frac{{{c^2}}}{3}$

$\Rightarrow$ $\quad x = \frac{C}{{\sqrt 3 }}\quad$ [using only positive sign]
Again, differentiating Eq. (ii) w.r.t. $x$, we get

$\frac{{{d^2}V}}{{d{x^2}}} = \frac{1}{4}( - 6x) = \frac{{ - 3}}{2}x < 0$

Therefore,${\left( {\frac{{{d^2}v}}{{d{x^2}}}} \right)_{{\rm{at}}\,x = \frac{c}{{\sqrt 3 }}}}$ $= - \frac{3}{2} \cdot \left( {\frac{c}{{\sqrt 3 }}} \right) < 0$

Thus, we see that volume $(V)$ is maximum at $x = \frac{C}{{\sqrt 3 }}$.

Therefore maximum volume of the box, ${(V)_{x = \frac{c}{{\sqrt 3 }}}} = \frac{1}{4}\left( {{c^2} \cdot \frac{c}{{\sqrt 3 }} - \frac{{{c^3}}}{{3\sqrt 3 }}} \right)$

$= \frac{1}{4} \cdot \frac{{\left( {3{c^3} - {c^3}} \right)}}{{3\sqrt 3 }} = \frac{1}{4} \cdot \frac{{2{c^3}}}{{3\sqrt 3 }}$

$= \frac{{{c^3}}}{{6\sqrt 3 }}{\rm{cu}}$ units