Question

A box with a square base and open top must have a volume of $13500{\text{cm}}^3$.

Find the dimensions of the box that minimize the amount of material used.

Sides of square base=___ $\text{cm}$

Height =____$\text{cm}$

• A. $30$
• B. $15$

Answer 1

Step-1. Give algebraic names to various measurements:

Let the square base of the box have sides of length $x\text{cm}$.

Thus, the length $l$ and width $w$ of the rectangular prism box are both $x\text{cm}$.

Also, let the height $h$ of the box be $y\text{cm}$.

Step-2: Determine an expression for the surface area of the required material:

The box has an open-top, and the surface area of the material required $S$ includes the lateral surface area of a rectangular prism and the area of the base.

The lateral surface area of a rectangular prism is $2\times h\times \left(l+w\right)$, in the present case that is, $2\times y\times \left(x+x\right)=4xy$.

The of the base is $l\times w$, in the present case that is, $x\times x=x^2$.

Thus, the surface area of the material required is $S=4xy+x^2$.

Step-3: Determine an expression for the surface area of the required material in terms of one variable only.

The volume $V{\text{cm}}^3$ of the rectangular prism box will be the product of its length $l=x\text{cm}$, width $w=x\text{cm}$, and height $h=y\text{cm}$.

Thus, $V=x^{2}y$. It is given that the volume $V{\text{cm}}^3$ of the box must be $13500{\text{cm}}^3$.

Substitute $V=13500$ in the equation $V=x^{2}y$ and then isolate $y$:

$\begin{array}{rcl}V& =& x^{2}y\\ & \Rightarrow & 13500=x^{2}y\\ & \Rightarrow & \frac{13500}{x^2}=y\end{array}$

Substitute, $\begin{array}{rcl}\frac{13500}{x^2}& =& y\end{array}$ in the equation $S=4xy+x^2$ to obtain $S$ in terms of $x$ alone:

$\begin{array}{rcl}S& =& 4xy+x^2\\ & =& 4x\left(\frac{13500}{x^2}\right)+x^2\\ & =& \frac{54000}{x}+x^2\end{array}$

Thus, the surface area of the material required is $\begin{array}{rcl}S& =& \frac{54000}{x}+x^2\end{array}$.

It is required to minimize $S$ with respect to $x$.

Step-4: Find the derivative of $S$ with respect to $x$.

$\begin{array}{rcl}S& =& \frac{54000}{x}+x^2\\ & \Rightarrow & \frac{dS}{dx}=\frac{d}{dx}\left(\frac{54000}{x}+x^2\right)\\ & \Rightarrow & \frac{dS}{dx}=54000\frac{d}{dx}\left(\frac{1}{x}\right)+\frac{d{x}^2}{dx}\\ & \Rightarrow & \frac{dS}{dx}=54000\frac{d\left(x^{-1}\right)}{dx}+2x\\ & \Rightarrow & \frac{dS}{dx}=54000\left(\left(-1\right)x^{-2}\right)+2x\\ & \Rightarrow & \frac{dS}{dx}=54000\left(\frac{-1}{x^2}\right)+2x\\ & \Rightarrow & \frac{dS}{dx}=\frac{-54000}{x^2}+2x\end{array}$

Thus, the derivative of $S$ is $\begin{array}{rcl}\frac{dS}{dx}& =& \frac{-54000}{x^2}+2x\end{array}$.

Step-5: Determine potential maxima/minima points of $S$

For maxima or minima, the derivative should be zero.

Solve $\begin{array}{rcl}\frac{dS}{dx}& =& 0\end{array}$ for $x$ to find possible points:

$\begin{array}{rcl}\frac{dS}{dx}& =& 0\\ & \Rightarrow & \frac{-54000}{x^2}+2x=0\\ & \Rightarrow & \frac{-54000+2x^3}{x^2}=0\\ & \Rightarrow & -54000+2x^3=0\\ & \Rightarrow & 2x^3=54000\\ & \Rightarrow & x^3=27000\\ & \Rightarrow & x=\sqrt[3]{27000}\\ & \Rightarrow & x=\sqrt[3]{30\times 30\times 30}\\ & \Rightarrow & x=30\end{array}$

Thus, $x=30$ is a possible minima of $S$.

Step-6: Find the second-order derivative of $S$ with respect to $x$.

$\begin{array}{rcl}\frac{dS}{dx}& =& \frac{-54000}{x^2}+2x\\ & \Rightarrow & \frac{d^{2}S}{d{x}^2}=\frac{d}{dx}\left(\frac{-54000}{x^2}+2x\right)\\ & \Rightarrow & \frac{d^{2}S}{d{x}^2}=-54000\frac{d}{dx}\left(\frac{1}{x^2}\right)+2\frac{dx}{dx}\\ & \Rightarrow & \frac{d^{2}S}{d{x}^2}=-54000\frac{d\left(x^{-2}\right)}{dx}+2\\ & \Rightarrow & \frac{d^{2}S}{d{x}^2}=-54000\left(\left(-2\right)\left(x^{-3}\right)\right)+2\\ & \Rightarrow & \frac{d^{2}S}{d{x}^2}=-54000\left(\frac{-2}{x^3}\right)+2\\ & \Rightarrow & \frac{d^{2}S}{d{x}^2}=\frac{108000}{x^3}+2\end{array}$

Thus, the second-order derivative of $S$ is $\begin{array}{rcl}\frac{d^{2}S}{d{x}^2}& =& \frac{108000}{x^3}+2\end{array}$.

Step-7: Use the second derivative test to confirm the minima of $S$.

Note that $\begin{array}{rcl}\frac{d^{2}S}{d{x}^2}& =& \frac{108000}{x^3}+2\end{array}$ will be positive for $x=30$.

Since $\begin{array}{rcl}\frac{dS}{dx}& =& 0\end{array}$, and $\begin{array}{rcl}\frac{d^{2}S}{d{x}^2}& >& 0\end{array}$ for $x=30$, by second derivative test $x=30$ is a minima of $S$.

Step-8: Find the value of the remaining variables at the minima of $S$.

Find the corresponding value of $y$ by substituting $x=30$ in the equation $\begin{array}{rcl}\frac{13500}{x^2}& =& y\end{array}$.

$\begin{array}{rcl}\frac{13500}{x^2}& =& y\\ & \Rightarrow & \frac{13500}{{\left(30\right)}^2}=y\\ & \Rightarrow & \frac{13500}{30\times 30}=y\\ & \Rightarrow & 15=y\end{array}$

Thus, $S$ is minimum for $x=30$, and $y=15$.

Hence, the surface area of the material required $\mathit{S}$ is minimum when the square base of the box has sides of length $\mathit{x}\mathbf{=}\mathbf{30}$, and the height of the box is $\mathit{y}\mathbf{=}\mathbf{15}$.