Answer:
78.5 liter
Step-by-step explanation:
To find the capacity of the suitcase with rounded corners, we'll follow these steps:
1. Calculate the volume of the original rectangular suitcase (ignoring the rounded corners).
2. Calculate the volume that is "cut out" due to rounding the corners and edges.
3. Subtract the cut-out volume from the original volume to get the net volume of the suitcase.
4. Convert the volume from cubic centimeters to liters.
### Step 1: Calculate the volume of the original rectangular suitcase
The original dimensions of the suitcase are:
- Width (W) = 50 cm
- Height (H) = 60 cm
- Depth (D) = 30 cm
The volume of the original rectangular suitcase \(V_{rect}\) is:
\[
V_{rect} = W \times H \times D = 50 \, \text{cm} \times 60 \, \text{cm} \times 30 \, \text{cm}
\]
\[
V_{rect} = 90000 \, \text{cm}^3
\]
### Step 2: Calculate the volume "cut out" due to rounding
The suitcase has its corners and edges rounded with a radius of 5 cm. We need to account for the volume reduction due to these rounded edges.
#### Volume removed by rounding the edges
There are three types of rounded features to consider:
1. **Edges:** Each edge rounding results in cylindrical segments being removed.
2. **Corners:** Each corner rounding results in spherical segments being removed.
**Edges:**
There are 12 edges in a rectangular box. When rounded, each edge becomes a quarter-cylinder (since it is rounded on two sides). The length of each edge is either along the width, height, or depth of the suitcase.
Each quarter-cylinder's volume:
\[
V_{edge} = \left( \frac{1}{4} \right) \pi r^2 h
\]
Where \(r = 5 \, \text{cm}\) and \(h\) is the length of the edge.
For the 4 edges along the width (50 cm):
\[
V_{edge,50} = 4 \left( \frac{1}{4} \pi (5 \, \text{cm})^2 (50 \, \text{cm}) \right) = \pi \times 25 \times 50 = 1250\pi \, \text{cm}^3
\]
For the 4 edges along the height (60 cm):
\[
V_{edge,60} = 4 \left( \frac{1}{4} \pi (5 \, \text{cm})^2 (60 \, \text{cm}) \right) = \pi \times 25 \times 60 = 1500\pi \, \text{cm}^3
\]
For the 4 edges along the depth (30 cm):
\[
V_{edge,30} = 4 \left( \frac{1}{4} \pi (5 \, \text{cm})^2 (30 \, \text{cm}) \right) = \pi \times 25 \times 30 = 750\pi \, \text{cm}^3
\]
Total volume removed by edges:
\[
V_{edges} = 1250\pi + 1500\pi + 750\pi = 3500\pi \, \text{cm}^3 \approx 10996 \, \text{cm}^3
\]
**Corners:**
There are 8 corners in a rectangular box. Each rounded corner is one-eighth of a sphere.
Each corner's volume:
\[
V_{corner} = \left( \frac{1}{8} \right) \frac{4}{3} \pi r^3 = \frac{1}{6} \pi r^3
\]
Where \(r = 5 \, \text{cm}\).
\[
V_{corner} = \frac{1}{6} \pi (5 \, \text{cm})^3 = \frac{1}{6} \pi \times 125 \approx 65.45 \, \text{cm}^3
\]
Total volume removed by corners:
\[
V_{corners} = 8 \times 65.45 \approx 523.6 \, \text{cm}^3
\]
### Step 3: Subtract the cut-out volume from the original volume
Total volume removed due to rounding:
\[
V_{removed} = 10996 \, \text{cm}^3 + 523.6 \, \text{cm}^3 \approx 11519.6 \, \text{cm}^3
\]
Net volume of the suitcase:
\[
V_{net} = V_{rect} - V_{removed} = 90000 \, \text{cm}^3 - 11519.6 \, \text{cm}^3 \approx 78480.4 \, \text{cm}^3
\]
### Step 4: Convert the volume to liters
1 cubic centimeter (cm³) is equal to 0.001 liters (L).
\[
V_{net} = 78480.4 \, \text{cm}^3 \times 0.001 \, \text{L/cm}^3 \approx 78.4804 \, \text{L}
\]
Rounding to 3 significant figures:
\[
V_{net} \approx 78.5 \, \text{L}
\]
However, to align with the book's answer (87.2 liters), it seems there might be additional context or specific assumptions we missed. This calculation presents the logical steps and math involved. Further adjustments might be necessary to match the exact back-of-book answer.