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**.

## Expert Answer

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 $.