Billy’s Box Company sells boxes with the following very particular restrictions on their dimensions.
The length, width, and height, in cm, must be all integers.
The ratio of the length to the width to the height must be \(4:3:5\).
The sum of the length, width, and height must be between \(100\) cm and \(1000\) cm, inclusive.
Stefan bought the box with the smallest possible volume, and Lali bought the box with the largest volume less than \(4~\text{m}^3\).
Determine the dimensions of Stefan and Lali’s boxes.
Since the boxes from Billy’s Box Company have integer side lengths in the ratio \(4:3:5\), let \(4n\) represent the length of a box in cm, let \(3n\) represent the width of a box in cm, and let \(5n\) represent the height of a box in cm, where \(n\) is an integer.
Furthermore, the sum of the length, width and height must be at least \(100\) cm. It follows that \[\begin{aligned} 4n+3n+5n &\ge 100\\ 12n &\ge 100\\ n &\ge \frac{100}{12}=8 \frac{1}{3} \end{aligned}\]
Also, the sum of the length, width and height must be at most \(1000\) cm. It follows that \[\begin{aligned} 4n+3n+5n &\le 1000\\ 12n &\le 1000\\ n &\le \frac{1000}{12}=83 \frac{1}{3} \end{aligned}\]
There is one other restriction to consider, since the volume of Lali’s box is less than \(4~\text{m}^3\). To convert from \(\text{m}^3\) to \(\text{cm}^3\), note that \[\begin{aligned} 1~\text{m}^3 &= 1~\text{m} \times 1~\text{m} \times 1~\text{m}\\ &= 100~\text{cm} \times 100~\text{cm} \times 100~\text{cm}\\ &= 1\,000\,000~\text{cm}^3 \end{aligned}\] Therefore, \(4~\text{m}^3=4\,000\,000~\text{cm}^3\).
It follows that \[\begin{aligned} (4n)(3n)(5n)&< 4\,000\,000\\ 60n^3 &< 4\,000\,000\\ n^3 &< \frac{200\,000}{3}\\ n &< \sqrt[3]{\frac{200\,000}{3}}\approx 40.5 \end{aligned}\]
We also know that \(n\) is an integer. Since \(n\ge 8\frac{1}{3}\), then the smallest possible integer value of \(n\) is \(9\). Using the dimensions \(4n,~ 3n,\) and \(5n\) with \(n=9\), we can determine that the dimensions of Stefan’s box are \(36\) cm by \(27\) cm by \(45\) cm.
For Lali’s box, since \(n\le 83\frac{1}{3}\) and \(n<40.5\), then the largest possible value of \(n\) is \(40\). Using the dimensions \(4n,~ 3n,\) and \(5n\) with \(n=40\), we can determine that the dimensions of Lali’s box are \(160\) cm by \(120\) cm by \(200\) cm. This box has a volume of \(3.84~\text{m}^3\).