Question

A box with a square base and open top must have a volume of $32000{\mathrm{cm}}^3$.

How do you find the dimensions of the box that minimize the amount of material used?

Answer 1

Minimizing the amount of material used to make an open-top square-base box:

Step-1: Finding the equations of volume and surface area.

Consider the volume of the box of base $a\times a$ and height $h$ as, $V=a^{2}h$.

The amount of surface area has to be minimized to minimize the material used.

The surface area is given by, $A=a^2+4ah$.

Substituting for $h$ from the first equation; we get, $A=a^2+4a\left(\frac{V}{a^2}\right)$

Volume is given as $32000c{m}^3$; substituting it, we get, $A=a^2+\frac{4\times 32000}{a}$:

$\Rightarrow A=a^2+\frac{128000}{a}$

Step-2: Finding the critical point by finding the first derivative and equating it to zero.

Differentiating $A$ with respect to $a$; we get,

$\frac{dA}{da}=2a-\frac{128000}{a^2}$

Equating the first derivative to zero; get the critical point as,

$\begin{array}{rcl}2a-\frac{128000}{a^2}& =& 0\\ & \Rightarrow & 2a=\frac{128000}{a^2}\\ & \Rightarrow & a^3=64000\\ & \Rightarrow & a=\sqrt[3]{64000}\\ & \Rightarrow & a=40\end{array}$

Step-3: Checking that the critical point gives the second derivative at that point as positive.

The second derivative of $A$ with respect to $a$, we get,

$$\begin{gathered} \frac{d^{2}A}{d{a}^2}=2-\left(\frac{2\times 128000}{a^3}\right) \\ \Rightarrow \frac{d^{2}A}{d{a}^2}=2+\frac{256000}{a^3} \end{gathered}$$

Now,

$\begin{array}{rcl}A\text{'}\text{'}\left(40\right)& =& 2+\frac{256000}{{40}^3}\\ & =& 2+\frac{256000}{64000}\\ & =& 2+4\\ & =& 6\end{array}$

At $a=40$, we get the second derivative as positive. Hence, the area is minimum.

Step-4: Finding the height of the box.

We get the height from the first equation as,

$\begin{array}{rcl}h& =& \frac{32000}{a^2}\\ & =& \frac{32000}{{40}^2}\\ & =& \frac{32000}{1600}\\ & =& 20\end{array}$

So, the dimensions of the box are: $40\mathrm{cm}\times 40\mathrm{cm}\times 20\mathrm{cm}$

Therefore, the required dimensions of the base of the box is $\mathbf{40}\mathbf{}\mathbf{cm}\mathbf{\times }\mathbf{40}\mathbf{}\mathbf{cm}$ and height, $\mathbf{20}\mathbf{}\mathbf{cm}$.