To solve for the value of $f$, we need to compare the volumes of the cone and the sphere and then solve the given equation involving $\pi/f$.

Step 1: Volume of the cone

The formula for the volume $V$ of a cone is:

$V_{\text{cone}} = \frac{1}{3} \pi r^2 h$

Where:

• $r$ is the radius of the base,
• $h$ is the height.

Given:

• $r = 2 \, \text{cm}$,
• $h = 3 \, \text{cm}$.

Thus, the volume of the cone is:

$V_{\text{cone}} = \frac{1}{3} \pi (2)^2 (3) = \frac{1}{3} \pi \times 4 \times 3 = 4 \pi \, \text{cm}^3$

Step 2: Volume of the sphere

The formula for the volume $V$ of a sphere is:

$V_{\text{sphere}} = \frac{4}{3} \pi r^3$

Where $r$ is the radius of the sphere.

Given that the sphere has a diameter of 3 cm, the radius is:

$r = \frac{3}{2} = 1.5 \, \text{cm}$

Thus, the volume of the sphere is:

$V_{\text{sphere}} = \frac{4}{3} \pi (1.5)^3 = \frac{4}{3} \pi \times 3.375 = 4.5 \pi \, \text{cm}^3$

Step 3: Relating the volumes

The new design contains $\frac{\pi}{f}$ centimeters more soap than the old design. Therefore, we can set up the equation:

$V_{\text{sphere}} = V_{\text{cone}} + \frac{\pi}{f}$

Substitute the values of $V_{\text{sphere}}$ and $V_{\text{cone}}$:

$4.5 \pi = 4 \pi + \frac{\pi}{f}$

Step 4: Solve for $f$

First, subtract $4 \pi$ from both sides:

$0.5 \pi = \frac{\pi}{f}$

Now, divide both sides by $\pi$ (since $\pi \neq 0$):

$0.5 = \frac{1}{f}$

Finally, solve for $f$:

$f = \frac{1}{0.5} = 2$

Final Answer:

The value of $f$ is $\boxed{2}$.

Would you like any additional details or clarifications?

Here are 5 related questions to consider:

1. How does the surface area of the sphere compare to the surface area of the cone?
2. What would be the volume of a cylinder with the same radius and height as the original cone?
3. How would the volumes change if the radius of both the cone and sphere were doubled?
4. How would the answer change if the cone had a different height but the same radius?
5. What is the formula for the surface area of a sphere, and how would it apply here?

Tip: When comparing volumes of different shapes, always make sure the units are consistent for accurate calculations.