 # A rectangular package to be sent by a postal service that has a maximum total length and perimeter (or girth) limit of 108 inches. A rectangular package is to be sent via this service. Calculate the dimensions of the package that covers the maximum volume. (Cross-sections may be assumed to be square) This question aims to learn the basic methodology for optimizing a mathematical function (maximizing or minimizing).

Critical points are the points where the value of a function is either maximum or minimum. To calculate the critical point(s), we equate the first derivative’s value to 0 and solve for the independent variable. We can use the second derivative test to find maxima/minima. If the value of $V’’(x)$ at the critical point is less than zero, then it’s a local maximum; otherwise, it’s a local minimum.

Let $x$, $y$, and $y$ be the dimensions of the rectangular box as shown in figure 1 below: Figure 1

Follow the steps to solve this question.

Step 1:  Calculate perimeter $P$:

$P = x + x + x + x + y$

$P = 4x + y$

Given that, $P = 108$

$y = 108 – 4x$

Step 2: Calculate Volume of the box $V(x)$:

$V(x, y) = x \cdot x \cdot y$

$V(x, y) = x^2 y$

Substituting value of $y$:

$V(x) = x^2 (108 – 4x)$

$V(x) = 108x^2-4x^3$

Step 3: Find the first and second derivatives:

$V’(x) = 2(108x)-3(4x^2)$

$V’(x) = 216x-12x^2$

$V’’(x) = 216 – 2(12x)$

$V’’(x) = 216 – 24x$

Step 4: At critical point(s), $V(‘x) = 0$:

$216x – 12x^2 = 0$

$x (216 – 12x) = 0$

This implies that either $x = 0$ or $216-12x = 0 \rightarrow x = \frac{216}{12} \rightarrow$ $x = 18$.

Step 5: Perform a Second derivative test:

Find $V’’(x)$ at $x = 18$ and $x = 0$,

$V’’(0) = 216 – 24(0) = 216 > 0 \rightarrow minima$

$V’’(18) = 216 – 24(18) = -216 < 0\rightarrow maxima$

Hence, volume $V$ is maximum at $x = 18$

Step 5: Final dimensions of the box:

$y = 108 – 4(18)$

$y = 36$

## Numerical Result

The maximum volume of the box is calculated as $18$ x $18$ x $36$ for the values of $x$, $y$ and $z$, respectively.

## Example

A rectangular package to be sent by a postal service that has a maximum total length and perimeter (or girth) limit of $54$ inches. A rectangular package is to be sent via this service. Calculate the dimensions of the package that covers the maximum volume (Cross-sections may be assumed to be square).

$P = 54 = 4x + y$

$y = 54 – 4x$

$V(x,y) = x^2 y = x^2 (54 – 4x) = 54x^2-4x^3$

$V’(x) = 108x – 12x^2 = 0$

This implies:

$x = 0 \ or\ x = 9$

$V’(x) = 108x – 12x^2 = 0$

Since:

$V’'(x) = 108 – 24x$

$V’'(9) = 108 – 24(9) = -108 > 0$

Maximum dimensions are $x = 9$ and $y = 108 – 4(9) = 72$.