Question

Pete runs an ice cream stand that also sells snow cones. The paper cones he usually have a diameter of 3 inches and a height of 4 inches, but his supplier is out of them. As a replacement, he purchases paper cones with a diameter of 4 inches and a height of 3 inches. How do the volumes of the original and replacement cones compare? (The original and replacement cones have the same volume; the original cone holds more than the replacement; the replacement cone holds more than the original; the replacement cone and original cone volumes cannot be found from the information)

Answer 1

To compare the volumes of the original and replacement cones, we can use the formula for the volume of a cone:

\[
V = \frac{1}{3} \pi r^2 h
\]

where \( r \) is the radius of the base, and \( h \) is the height of the cone.

**Original Cone:**
- Diameter = 3 inches, so radius \( r = \frac{3}{2} = 1.5 \) inches
- Height \( h = 4 \) inches

Now we can calculate the volume of the original cone:

\[
V_{\text{original}} = \frac{1}{3} \pi (1.5)^2 (4)
\]
\[
V_{\text{original}} = \frac{1}{3} \pi (2.25) (4)
\]
\[
V_{\text{original}} = \frac{1}{3} \pi (9)
\]
\[
V_{\text{original}} = 3 \pi \text{ cubic inches}
\]

**Replacement Cone:**
- Diameter = 4 inches, so radius \( r = \frac{4}{2} = 2 \) inches
- Height \( h = 3 \) inches

Now we can calculate the volume of the replacement cone:

\[
V_{\text{replacement}} = \frac{1}{3} \pi (2)^2 (3)
\]
\[
V_{\text{replacement}} = \frac{1}{3} \pi (4) (3)
\]
\[
V_{\text{replacement}} = \frac{1}{3} \pi (12)
\]
\[
V_{\text{replacement}} = 4 \pi \text{ cubic inches}
\]

**Comparing the Volumes:**
- \( V_{\text{original}} = 3\pi \)
- \( V_{\text{replacement}} = 4\pi \)

Since \( 4\pi > 3\pi \):

The replacement cone holds more than the original cone.

Thus, the answer is:
**The replacement cone holds more than the original.**