Chapter 4.6, Problem 1E

To determine

The dimensions of the box for the least material.

Answer to Problem 1E

The length, breadth and height of the box for the least material are $2\text{ft}$ , $2\text{ft}$ and $1\text{ft}$ respectively.

Explanation of Solution

Given:

The volume of box is $4{\text{ft}}^{\text{3}}$ .

Calculation:

Let the length and height of box as $x$ and $h$ respectively.

The volume of box is calculated as follows:

  $\begin{array}{l}V=x^{2}h\\ h=\frac{4}{x^2}\end{array}$

The area of material is calculated as follows:

  $\begin{array}{c}A=x^2+4xh\\ =x^2+4x\left(\frac{4}{x^2}\right)\\ =x^2+\frac{16}{x}\end{array}$

For critical points, derivate the above area with respect to x and equate to zero as follows:

  $\begin{array}{c}\frac{dA}{dx}=\frac{d}{dx}\left(x^2+\frac{16}{x}\right)\\ 0=\left(2x-\frac{16}{x^2}\right)\\ 0=\left(2x^3-16\right)\\ x=2\end{array}$

Again differentiate the volume for maximum and minimum condition at critical point as follows:

  $\begin{array}{c}\frac{d^{2}A}{d{x}^2}=\frac{d^2}{d{x}^2}\left(\left(x^2+\frac{16}{x}\right)\right)\\ =\frac{d}{dx}\left(2x-\frac{16}{x^2}\right)\\ =\left(2+\frac{32}{x^3}\right)\end{array}$

This clearly shows that $\frac{d^{2}A}{d{x}^2}$ has positive value which signifies the minimum condition.

The height of least material of box is calculated as follows:

  $\begin{array}{c}h=\frac{4}{x^2}\\ =\frac{4}{2^2}\\ =1\text{ft}\end{array}$

Thus, the length, breadth and height of the box for the least material are $2\text{ft}$ , $2\text{ft}$ and $1\text{ft}$ respectively.