Question

The box with a square base is to have an open top. The area of the material for making the box is $192\;{\text{sq}}{\text{.}}\;{\text{cm}}$. What should be its dimensions in order that the volume is as large as possible?

Answer 1

Hint:
We know that the base of the box is square, so two dimensions of the box will be the same. Use the first order derivative test to find the critical point. We need to apply the second order derivative test to find the point of maximum such that the volume of the box is largest.

Complete step by step solution:
We know that the base of the box is in square shape. So, the length and breadth of the box will be the same.
 Let us consider that the length of the box is $x\;{\text{cm}}$, breadth is $x\;{\text{cm}}$ and height is $y\;{\text{cm}}$. The surface area of the box with an open top is:
${\text{Area}}\;{\text{of}}\;{\text{box}} = \left( {{\text{area}}\;{\text{of}}\;{\text{base}}} \right) + \left( {{\text{area}}\;{\text{of}}\;{\text{4}}\;{\text{sides}}} \right)$
Substitute all the values and solve for the height of the box.
$ \Rightarrow 192 = \left( {x \cdot x} \right) + 4\left( {xy} \right)$
$ \Rightarrow 192 = {x^2} + 4xy$
$ \Rightarrow 4xy = 192 - {x^2}$
$ \Rightarrow y = \dfrac{{192 - {x^2}}}{{4x}}\;\;...\left( 1 \right)$
We know that the formula of the volume $V$ of an open box is the product of its base area and height. So, we have
$ \Rightarrow V = {x^2}y$
$ \Rightarrow V = {x^2}\left( {\dfrac{{192 - {x^2}}}{{4x}}} \right)$
$ \Rightarrow V = \dfrac{{192x - {x^3}}}{4}$
For the maximum volume of the box first, we need to find the critical point and then use the second order derivative test for the maximum point.
Differentiate the volume with respect to $x$.
$\Rightarrow \dfrac{{dV}}{{dx}} = \dfrac{1}{4}\dfrac{d}{{dx}}\left( {192x - {x^3}} \right)$
$ \Rightarrow \dfrac{{dV}}{{dx}} = \dfrac{{\left( {192 - 3{x^2}} \right)}}{4}$
Now, equate the first order derivative to $0$ for the critical point.
$\Rightarrow \dfrac{{dV}}{{dx}} = 0$
$\Rightarrow \dfrac{{\left( {192 - 3{x^2}} \right)}}{4} = 0$
$\Rightarrow 192 - 3{x^2} = 0$
$\Rightarrow {x^2} = 64$
$\Rightarrow x = \pm 8$
As we know that the length cannot be negative. So, we get $x = 8\;{\text{cm}}$.
Again differentiate the function with respect to $x$.
$\dfrac{{{d^2}V}}{{d{x^2}}} = - \dfrac{3}{2}x$
At $x = 8\;{\text{cm}}$:
$\Rightarrow {\left( {\dfrac{{{d^2}V}}{{d{x^2}}}} \right)_{x = 8}} = - \dfrac{3}{2}\left( 8 \right)$
$\Rightarrow {\left( {\dfrac{{{d^2}V}}{{d{x^2}}}} \right)_{x = 8}} = - 12 < 0$
The volume of the box is maximum at $x = 8\;{\text{cm}}$ by the second derivative test as the value of the second derivative at this point is negative.
Now, substitute the value in equation $\left( 1 \right)$ to find the height of the box.
$\Rightarrow y = \dfrac{{192 - {{\left( 8 \right)}^2}}}{{4\left( 8 \right)}}$
$\Rightarrow y = 4$
The required measurements of the box are $x = 8\;{\text{cm}}$ and $y = 4\;{\text{cm}}$.
The possible maximum volume of box of an open box is:
$
 \Rightarrow V = {\left( {8\;{\text{cm}}} \right)^2}\left( {4\;{\text{cm}}} \right) \\
 \Rightarrow V = 256\;{\text{c}}{{\text{m}}^3} \\
 $

Therefore, the required dimensions of the box are $8\;{\text{cm}}$, $8\;{\text{cm}}$ and $4\;{\text{cm}}$ such that box has maximum volume.

Note:
In this question, find the point of local maximum for the largest volume of the box. For any point $x = c$, use a second derivative test to check whether the function attains maximum value or minimum value. If $f''\left( c \right) < 0$ then the function has maximum value and if $f''\left( c \right) > 0$ then the function has minimum value.