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Given: a cardboard box that is open and square in shape has $c^2$ area

To show: $\frac{c^3}{6\sqrt3}$ cubic units is the maximum volume of the box.

Explanation:

Take the side of the square be x cm and

Take the height the box be y cm.

So, the total area of the cardboard used is

A = area of square base + 4x area of rectangle

$\Rightarrow A = x\textsuperscript{2}+4xy$

But it is given this is equal to $c\textsuperscript{2}$, hence

$\\{c\textsuperscript{2} = x\textsuperscript{2}+4xy}\\ { \Rightarrow 4xy = c\textsuperscript{2}-x\textsuperscript{2}}\\$

$\Rightarrow \mathrm{y}=\frac{\mathrm{c}^{2}-\mathrm{x}^{2}}{4 \mathrm{x}} \ldots \ldots(i)$

According to the given condition the area of the square base will be

V = base × height

Since the base is square, the volume is

$V = x\textsuperscript{2}y \ldots \ldots (ii)$

Then putting the values of equation (i) in equation (ii), we get

$\\ \mathrm{V}=\mathrm{x}^{2}\left(\frac{\mathrm{c}^{2}-\mathrm{x}^{2}}{4 \mathrm{x}}\right) \\ \mathrm{V}=\frac{\mathrm{x}}{4}\left(\mathrm{c}^{2}-\mathrm{x}^{2}\right) \\ \mathrm{V}=\frac{1}{4}\left(\mathrm{xc}^{2}-\mathrm{x}^{3}\right)$
Calculation of the first derivative of the equation,

$\mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\frac{1}{4}\left(\mathrm{xc}^{2}-\mathrm{x}^{3}\right)\right)}{\mathrm{dx}}$
Removing all the constant terms
$\mathrm{V}^{\prime}=\frac{1}{4} \frac{\mathrm{d}\left(\mathrm{xc}^{2}-\mathrm{x}^{3}\right)}{\mathrm{dx}}$
Using the sum rule of differentiation, we get
$\mathrm{V}^{\prime}=\frac{1}{4}\left[\frac{\mathrm{d}\left(\mathrm{xc}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}\right]$

Removing all the constant terms, we get
$\mathrm{V}^{\prime}=\frac{1}{4}\left[\mathrm{c}^{2} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{d} \mathrm{x}}\right]$
After differentiating the equation, we get
$\mathrm{V}^{\prime}=\frac{1}{4}\left[\mathrm{c}^{2}-3 \mathrm{x}^{2}\right] \ldots \ldots (iii)$
We need to calculate the second derivative to find out the maximum value of x , so for that let $\mathrm{V}^{\prime}=0, so$ equating above equation with 0, we get
$\frac{1}{4}\left[c^{2}-3 x^{2}\right]=0$

$\\ \Rightarrow c^{2}-3 x^{2}=0 \\\\\Rightarrow 3 x^{2}=c^{2} \\\\\Rightarrow \mathrm{x}^{2}=\frac{\mathrm{c}^{2}}{3} \\\\\Rightarrow \mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}$
Differentiating equation (iii) again with respect to x, we get
$\Rightarrow \mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(\frac{1}{4}\left[\mathrm{c}^{2}-3 \mathrm{x}^{2}\right]\right)}{\mathrm{dx}}$
Removing all the constant terms results into
$\Rightarrow \mathrm{V}^{\prime \prime}=\frac{1}{4} \frac{\mathrm{d}\left(\mathrm{c}^{2}-3 \mathrm{x}^{2}\right)}{\mathrm{dx}}$

Using the differentiation rule of sum, we get
$\\\Rightarrow \mathrm{V}^{\prime \prime}=\frac{1}{4}\left[\frac{\mathrm{d}\left(\mathrm{c}^{2}\right)}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}\right] \\\Rightarrow \mathrm{V}^{\prime \prime}=\frac{1}{4}[0-3(2 \mathrm{x})] \\\Rightarrow \mathrm{V}^{\prime \prime}=\frac{-6 \mathrm{x}}{4} \\\Rightarrow \mathrm{V}^{\prime \prime}=\frac{-3 \mathrm{x}}{2}$
$x=\frac{c}{\sqrt{3}}$, the above equation becomes,
$\\\Rightarrow\left(\mathrm{V}^{\prime \prime}\right)_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{-3\left(\frac{\mathrm{c}}{\sqrt{3}}\right)}{2} \\\\\Rightarrow\left(\mathrm{V}^{\prime \prime}\right)_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{-3 \mathrm{c}}{2 \sqrt{3}}<0$

Thus, the volume (V) is maximum at $x=\frac{c}{\sqrt{3}}$

∴ The box has a maximum value of

$\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{1}{4}\left(\left(\frac{\mathrm{c}}{\sqrt{3}}\right) \mathrm{c}^{2}-\left(\frac{\mathrm{c}}{\sqrt{3}}\right)^{3}\right) \\\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{1}{4}\left(\left(\frac{\mathrm{c}^{3}}{\sqrt{3}}\right)-\frac{\mathrm{c}^{3}}{3 \sqrt{3}}\right) \\\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{1}{4}\left(\frac{\mathrm{c}^{3}}{\sqrt{3}}\left(1-\frac{1}{3}\right)\right) \\\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{1}{4}\left(\frac{\mathrm{c}^{3}}{\sqrt{3}}\left(\frac{2}{3}\right)\right) \\\\ \mathrm{V}_{\mathrm{x}=\frac{\mathrm{c}}{\sqrt{3}}}=\frac{\mathrm{c}^{3}}{6 \sqrt{3}} \mathrm{units}$

So, the box has a maximum value of is $\frac{\mathrm{c}^{3}}{6 \sqrt{3}}$ cubic units.

Hence, proved.

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