Question

\(\quad\) A box has a length of \((52-2 x)\) inches, a width of \((42-2 x)\) inches, and a height of \(x\) inches. Find the volume when \(x=3, x=7\), and \(x=9\) inches. Which \(x\) -value gives the greatest volume?

Short answer

The \(x\)-value of \(3\) inches gives the greatest volume.

Step-by-step solution

Understand the Problem

The problem is to find the volume of a box whose dimensions are given in terms of variable \(x\). The formula for calculating volume is \(Volume = length \times width \times height\).

Substitute \(x\) values in Volume formula

Substitute \(x=3, x=7, x=9\) into the equation \(V = (52 - 2x) \times (42 - 2x) \times x\) separately to determine the respective volumes. This yields \(V_{3} = (52-2*3)*(42-2*3)*3\), \(V_{7} = (52-2*7)*(42-2*7)*7\) and \(V_{9} = (52-2*9)*(42-2*9)*9\).

Calculate the volume for each \(x\) - value

Now, calculate the volume for the respective \(x\) - values using the obtained equations. This results in \(V_{3} = 15804, V_{7} = 11907, V_{9} = 7560\).

Compare Volumes

Compare the volume values obtained in the previous step. The greatest among \(V_{3}, V_{7}, V_{9}\) will indicate which \(x\) -value provides the maximum volume.

Results

Based on comparison, \(V_{3} = 15804\) is the greatest preceded by \(V_{7} = 11907\) and \(V_{9} = 7560\). So, \(x = 3\) inches yields the largest volume.

Key concepts

Understanding Length in Box Dimensions

The concept of length is crucial when dealing with the dimensions of a box. Length refers to how far the box extends in one direction. In our exercise, the length of the box changes based on the variable \(x\). It is expressed as \((52 - 2x)\) inches. As \(x\) changes, this expression shows how the length shortens. Understanding this helps in determining how the total volume of the box varies with different values of \(x\).
\(\)
Here’s a simple breakdown:

• When \(x = 3\), the length is \(52 - 2 \times 3 = 46\) inches.
• For \(x = 7\), the length is \(52 - 2 \times 7 = 38\) inches.
• And when \(x = 9\), it decreases to \(52 - 2 \times 9 = 34\) inches.

These calculations show that increasing \(x\) decreases the length. This affects the overall volume of the box, as a smaller length typically results in a smaller volume for a fixed height and width.

Exploring Width in Box Measurements

Width is another dimension that influences the overall volume of a box. It's how wide the box is and is given as \((42 - 2x)\) inches in this case. Similar to length, the width also alters as \(x\) varies. This change must be considered to fully grasp how volume changes.
\(\)
Here’s how width is calculated with different \(x\) values:

• For \(x = 3\), the width is \(42 - 2 \times 3 = 36\) inches.
• When \(x = 7\), it is \(42 - 2 \times 7 = 28\) inches.
• When \(x = 9\), it further reduces to \(42 - 2 \times 9 = 24\) inches.

As observed, the width decreases as \(x\) increases, similar to the length. A narrower width combined with a shorter length directly impacts the potential volume, making these dimensions important to consider critically when calculating the volume.

Defining the Role of Height in the Volume Formula

Height is the vertical dimension of our box and also varies with \(x\) in this particular problem. In this exercise, height is simply \(x\) inches.
\(\)
The role of height is straightforward but vital because it directly multiplies with length and width in the volume formula: \[ \text{Volume} = (\text{length}) \times (\text{width}) \times (\text{height}) \]
\(\)
Let's consider the specifics:

• When \(x = 3\), the height is \(3\) inches.
• When \(x = 7\), it becomes \(7\) inches.
• At \(x = 9\), the height is \(9\) inches.

The height increases with \(x\), contrasting the behavior of length and width. This means while the box’s base might shrink with larger \(x\), a greater height may partially compensate by increasing the volume. However, finding the optimal \(x\) means balancing these interactions to find which set of dimensions gives the maximum volume, as we discovered \(x = 3\) delivers the largest volume.