Question

An open box with a square base needs to be made out of a given quantity of cardboard of area ${{c}^{2}}$ units. Show that the maximum volume of the box is $\dfrac{{{c}^{3}}}{6\sqrt{3}}$ cubic units.

Answer 1

Hint: We solve this problem by first finding the area of the box using the formula $2lb+2bh+2lh$ and then equating the obtained value with the area of cardboard. Then we can find the height in terms of base length and c. Then we find the volume of the box using the formula $lbh$ and substitute the value of height as obtained before. Then we differentiate volume and equate it with zero. By that we can find the value of the length of the square base. Then by substituting it in the volume we can find the maximum value of the volume possible.

Complete step by step answer:
We were given that an open box is to be made from a cardboard of area ${{c}^{2}}$ units.
Then, however we make the changes the total area of cardboard is constant.
Let us consider the length of the side of the square base as $a$ and let the height of the box be $h$.

Now, let us find the area of the box made from the above dimensions.
Let us consider the formula for area of the cuboid of length l, breadth b and height h.,
$\Rightarrow \text{Area of cuboid }=2lb+2bh+2lh$
But as our box is open on the top, we can subtract the area of the wall on the top.
$\begin{align}
  & \Rightarrow \text{Area of box }=\text{ 2}\times \left( a\times a \right)+2\times \left( a\times h \right)+2\times \left( a\times h \right)-\left( a\times a \right) \\
 & \Rightarrow \text{Area of box }=\text{ 2}{{a}^{2}}+2ah+2ah-{{a}^{2}} \\
 & \Rightarrow \text{Area of box }=\text{ }{{a}^{2}}+4ah \\
\end{align}$
So, now let us equate it to the area of the cardboard.
$\begin{align}
  & \Rightarrow {{a}^{2}}+4ah={{c}^{2}} \\
 & \Rightarrow 4ah={{c}^{2}}-{{a}^{2}} \\
 & \Rightarrow h=\dfrac{{{c}^{2}}-{{a}^{2}}}{4a}................\left( 1 \right) \\
\end{align}$
Now let us find the volume of the box.
Let us consider the formula for the volume of a cuboid of length l, breadth b and height h.
$\Rightarrow \text{Volume of cuboid }=lbh$
Here in our problem the length and breadth of the box is a unit and height is h units. Then using the above formula volume of the box is
$\Rightarrow \text{Volume of box }=V=a\times a\times h$
Using equation (1) and substituting in it, we get
$\begin{align}
  & \Rightarrow V={{a}^{2}}h \\
 & \Rightarrow V={{a}^{2}}\left( \dfrac{{{c}^{2}}-{{a}^{2}}}{4a} \right) \\
 & \Rightarrow V=a\left( \dfrac{{{c}^{2}}-{{a}^{2}}}{4} \right) \\
 & \Rightarrow V=\dfrac{a{{c}^{2}}-{{a}^{3}}}{4} \\
\end{align}$
Now let us discuss the concept of optimization. A function $f\left( x \right)$ is said to achieve extremum that is maximum or minimum if it satisfies the condition,
${f}'\left( x \right)=0$
As we need to find the maximum value of volume, let us differentiate the above volume with respective to a and equate it to zero.
$\begin{align}
  & \Rightarrow \dfrac{dV}{da}=\dfrac{d\left( \dfrac{a{{c}^{2}}-{{a}^{3}}}{4} \right)}{da} \\
 & \Rightarrow \dfrac{dV}{da}=\dfrac{1}{4}\dfrac{d\left( a{{c}^{2}}-{{a}^{3}} \right)}{da} \\
 & \Rightarrow \dfrac{dV}{da}=\dfrac{1}{4}\left( {{c}^{2}}-3{{a}^{2}} \right) \\
 & \Rightarrow \dfrac{1}{4}\left( {{c}^{2}}-3{{a}^{2}} \right)=0 \\
 & \Rightarrow {{c}^{2}}-3{{a}^{2}}=0 \\
 & \Rightarrow 3{{a}^{2}}={{c}^{2}} \\
 & \Rightarrow {{a}^{2}}=\dfrac{{{c}^{2}}}{3} \\
 & \Rightarrow a=\dfrac{c}{\sqrt{3}} \\
\end{align}$
Now, let us substitute the value of a in the volume of the box.
$\begin{align}
  & \Rightarrow V=\dfrac{a{{c}^{2}}-{{a}^{3}}}{4} \\
 & \Rightarrow V=\dfrac{\left( \dfrac{c}{\sqrt{3}} \right){{c}^{2}}-{{\left( \dfrac{c}{\sqrt{3}} \right)}^{3}}}{4} \\
 & \Rightarrow V=\dfrac{\dfrac{{{c}^{3}}}{\sqrt{3}}-\dfrac{{{c}^{3}}}{3\sqrt{3}}}{4} \\
 & \Rightarrow V=\dfrac{\dfrac{2{{c}^{3}}}{3\sqrt{3}}}{4} \\
 & \Rightarrow V=\dfrac{2{{c}^{3}}}{12\sqrt{3}} \\
 & \Rightarrow V=\dfrac{{{c}^{3}}}{6\sqrt{3}} \\
\end{align}$
Hence the maximum volume of the box is $\dfrac{{{c}^{3}}}{6\sqrt{3}}$ square units.

Note: While solving the problem one can make a mistake of not subtracting the area of the top wall from the formula $2lb+2bh+2lh$. If we do not subtract it, it will give the area of the box with top closed not that of an open box. Similarly, another mistake is possible that one might confuse by subtracting the area of the top wall from the volume $lbh$. But we should not do that because the volume of the box is the same despite the fact that the box is open or closed.