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

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

The correct answer is Option (4) → $\frac{c^3}{6\sqrt{3}}$ cu units ##

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

$∴$ Area of the metal used $= x^2 + 4xy$

$\Rightarrow x^2 + 4xy = c^2 \quad \text{[given]}$

$\Rightarrow y = \frac{c^2 - x^2}{4x} \quad \dots \text{(i)}$

Now, volume of the box $(V) = x^2y \quad [∵v = l \times b \times h]$

$\Rightarrow V = x^2 \cdot \left( \frac{c^2 - x^2}{4x} \right) \quad \text{[From Eq. (i)]}$

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

On differentiating both sides w.r.t. $x$, we get

$\frac{dV}{dx} = \frac{1}{4}(c^2 - 3x^2) \quad \dots \text{(ii)}$

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

$\Rightarrow x^2 = \frac{c^2}{3}$

$\Rightarrow x = \frac{c}{\sqrt{3}} \quad \text{[on using positive sign]}$

Again, differentiating Eq. (ii) w.r.t. $x$, we get

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

$∴\left( \frac{d^2V}{dx^2} \right)_{\text{at } x = \frac{c}{\sqrt{3}}} = -\frac{3}{2} \cdot \left( \frac{c}{\sqrt{3}} \right) < 0$

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

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

$= \frac{1}{4} \frac{(3c^3 - c^3)}{3\sqrt{3}} = \frac{1}{4} \cdot \frac{2c^3}{3\sqrt{3}} = \frac{c^3}{6\sqrt{3}} \text{ cu units}$