Question

Ravi has a briefcase with dimensions of two by 1 ½ by 1 ½ feet. He wishes to place several notebooks, each eight by nine by one inch, into the briefcase. What is the largest number of notebooks the briefcase will hold?

Hint:

Convert every length to the same unit first. Then divide the briefcase volume by the notebook volume - this gives the highest possible count. If you can then find one arrangement where every notebook edge divides a briefcase edge with no remainder, that arrangement actually reaches this maximum, so you do not need to test any other orientation.

Answer 1

Step 1: Understanding the Question:
This is a 3D packing problem. We need to find the maximum number of small boxes (notebooks) that can fit inside a larger box (briefcase). We must ensure the units are consistent before calculating.
Step 2: Key Formula or Approach:
1. Convert all dimensions to the same unit (inches).
2. Calculate the volume of the briefcase and the notebook.
3. To find the maximum number, it's best to check how many notebooks fit along each dimension (length, width, height) rather than just dividing volumes, as this accounts for packing orientation.
Step 3: Detailed Explanation:
Part 1: Convert dimensions to inches
Given that 1 foot = 12 inches.
Briefcase dimensions: 2 feet \(\times\) 1.5 feet \(\times\) 1.5 feet.
Length = \(2 \times 12 = 24\) inches.
Width = \(1.5 \times 12 = 18\) inches.
Height = \(1.5 \times 12 = 18\) inches.
So, the briefcase is 24" \(\times\) 18" \(\times\) 18".
Notebook dimensions: 9 inches \(\times\) 8 inches \(\times\) 1 inch.
Part 2: Check packing orientation
We need to find the most efficient way to pack the 9x8x1 notebooks into the 24x18x18 briefcase.
Orientation 1:
- Along the 24-inch dimension, we can fit \(\lfloor \frac{24}{9} \rfloor = 2\) notebooks or \(\lfloor \frac{24}{8} \rfloor = 3\) notebooks. Let's try fitting the 8-inch side. - Number along Length (24"): \(\frac{24}{8} = 3\) notebooks.
- Number along Width (18"): \(\frac{18}{9} = 2\) notebooks.
- Number along Height (18"): \(\frac{18}{1} = 18\) notebooks.
Total notebooks in this orientation = \(3 \times 2 \times 18 = 108\) notebooks.
Orientation 2:
- Let's try fitting the 9-inch side along the 24-inch dimension. - Number along Length (24"): \(\lfloor \frac{24}{9} \rfloor = 2\) notebooks.
- Number along Width (18"): \(\lfloor \frac{18}{8} \rfloor = 2\) notebooks.
- Number along Height (18"): \(\frac{18}{1} = 18\) notebooks.
Total notebooks in this orientation = \(2 \times 2 \times 18 = 72\) notebooks.
Comparing the orientations, the first one yields the largest number.
Step 4: Final Answer:
The largest number of notebooks the briefcase can hold is 108.

Answer 2

Concept:

• For non-overlapping boxes packed inside a larger box, the total volume of the smaller boxes can never exceed the volume of the container. Dividing the two volumes gives an upper bound on how many can possibly fit.
• If an arrangement can be found where every notebook edge divides some briefcase edge with zero remainder, that arrangement reaches the volume bound exactly - proving it is the true maximum, not just the best of a few guesses.

Step 1: Convert every length to the same unit.
Since 1 foot = 12 inches, the briefcase (2 ft x 1 1/2 ft x 1 1/2 ft) becomes 24 in x 18 in x 18 in. The notebook stays at 9 in x 8 in x 1 in.

Step 2: Find the volume-based upper bound.
Volume of briefcase: $24 \times 18 \times 18 = 7776 \text{ in}^3$.
Volume of one notebook: $9 \times 8 \times 1 = 72 \text{ in}^3$.
Upper bound on the number of notebooks: $7776 \div 72 = 108$. No packing can ever exceed this, since notebooks cannot overlap.

Step 3: Check whether an exact-fit arrangement reaches this bound.
Test whether each notebook edge divides some briefcase edge with no leftover:
$24 \div 8 = 3$ (exact)
$18 \div 9 = 2$ (exact)
$18 \div 1 = 18$ (exact)
All three divisions come out exact, with zero remainder in every direction.

Step 4: Conclude.
Stacking 3 notebooks (8-inch edge) along the 24-inch length, 2 notebooks (9-inch edge) along one 18-inch side, and 18 notebooks (1-inch edge) along the other 18-inch side fills the briefcase with zero wasted space. This arrangement reaches the volume upper bound of 108 exactly, proving 108 is the true maximum.

Final Answer: 108 notebooks