Question

8inches tall 4iches diameter if both dimensions are reduced by one half inch by what percentage is the volume of the container reduced round to nearest percentage

Answer 1

To find the percentage reduction in the volume of the container when the dimensions are each reduced by half an inch, we will follow these steps:

1. **Calculate the Original Volume**:
The container is a cylinder, with the volume \( V \) calculated using the formula:
\[
V = \pi r^2 h
\]
where \( r \) is the radius and \( h \) is the height.

Given:
- Height \( h = 8 \) inches
- Diameter \( d = 4 \) inches, thus the radius \( r = \frac{d}{2} = \frac{4}{2} = 2 \) inches.

Now, substituting the values into the volume formula:
\[
V_{\text{original}} = \pi (2)^2 (8) = \pi (4)(8) = 32\pi \, \text{cubic inches}.
\]

2. **Calculate the New Volume**:
When both dimensions are reduced by half an inch:
- New height \( h_{\text{new}} = 8 - 0.5 = 7.5 \) inches
- New radius \( r_{\text{new}} = 2 - 0.5 = 1.5 \) inches.

Now calculate the new volume:
\[
V_{\text{new}} = \pi (1.5)^2 (7.5) = \pi (2.25)(7.5) = 16.875\pi \, \text{cubic inches}.
\]

3. **Calculate the Volume Reduction**:
The volume reduction is:
\[
\Delta V = V_{\text{original}} - V_{\text{new}} = 32\pi - 16.875\pi = (32 - 16.875)\pi = 15.125\pi \, \text{cubic inches}.
\]

4. **Calculate the Percentage Reduction**:
The percentage reduction in volume can be calculated as:
\[
\text{Percentage Reduction} = \left( \frac{\Delta V}{V_{\text{original}}} \right) \times 100\% = \left( \frac{15.125\pi}{32\pi} \right) \times 100\% = \left( \frac{15.125}{32} \right) \times 100\%.
\]

\[
\text{Percentage Reduction} \approx \left( 0.47265625 \right) \times 100\% \approx 47.27\% \approx 47\% \text{ (rounded to the nearest percentage)}.
\]

Thus, the percentage reduction in the volume of the container is approximately **47%**.