Chapter 1.1, Problem 33PPSE

Solution

To identify the pattern in which the dimensions of the box changes and to find the volume of the extra-large box which is one size larger than the third box.

From the given pattern of boxes, it is observed that the width of the boxes remains the same: $16\text{in}$ . Also, it is observed that the height and the length of the boxes increase by $4$ inches from that of the previous one and are all equal throughout the pattern.

Since the extra-large box is formed out of the third box by expanding it by $1$ in size, the dimensions of the extra-large box is found to be:

  $\begin{array}{c}l=24\text{in}\\ w=16\text{in}\\ h=24\text{in}\end{array}$ .

Then, the volume $V$ of the extra-large box is found using the formula $V=lwh$ :

  $V=9216{\text{in}}^3$

Given:

Given the boxes:

  

Also, it is given that the extra-large box is one size greater than the third one.

Concept Used:

Volume of a rectangular prism:

Suppose that $l$ represents the length, $w$ represents the width and $h$ represents the height of the prism. Then, the volume $V$ of the prism is given as:

  $V=lwh$ .

Calculation:

Determine the pattern of change in the dimension of boxes:

From the given pattern of boxes, observe that the width of the boxes remains the same: $16\text{in}$ .

Now, both the height and the length of the boxes increase by $4$ inches from that of the previous one and are all equal throughout the pattern.

Thus, the width $w$ of the extra-large box will be $16\text{in}$ .

Also, the length $l$ and the height $h$ of the extra-large box are all the same: $4$ inches more than those of the third one.

Thus, the length $l$ and the height $h$ of the extra-large box:

  $\begin{array}{l}l=20+4\text{in}\\ h=20+4\text{in}\end{array}$ .

That is, the length $l$ , the height $h$ and the width $w$ of the extra-large box:

  $\begin{array}{c}l=24\text{in}\\ w=16\text{in}\\ h=24\text{in}\end{array}$ .

Find the volume of the extra-large box:

The dimensions of the extra-large box is found:

  $\begin{array}{c}l=24\text{in}\\ w=16\text{in}\\ h=24\text{in}\end{array}$ .

Then, the volume $V$ of the extra-large box:

  $\begin{array}{l}V=lwh{\text{in}}^3\\ V=24\cdot 16\cdot 24{\text{in}}^3\\ V=9216{\text{in}}^3\end{array}$ .

Thus, the volume of the extra-large box is $9216{\text{in}}^3$ .

Conclusion:

From the given pattern of boxes, it is observed that the width of the boxes remains the same: $16\text{in}$ . Also, it is observed that the height and the length of the boxes increase by $4$ inches from that of the previous one and are all equal throughout the pattern.

Since the extra-large box is formed out of the third box by expanding it by $1$ in size, the dimensions of the extra-large box is found to be:

  $\begin{array}{c}l=24\text{in}\\ w=16\text{in}\\ h=24\text{in}\end{array}$ .

Then, the volume $V$ of the extra-large box is found using the formula $V=lwh$ :

  $V=9216{\text{in}}^3$