A cube with edge length 1.6' contains a sphere that touches all 6 sides of the box. Find the difference between the volume of the cube and the sphere, rounded to the nearest tenth.

1. Problem Description

A cube with edge length 1.6' contains a sphere that touches all 6 sides of the box. Find the difference between the volume of the cube and the sphere, rounded to the nearest tenth.

2. Solution Steps

First, we find the volume of the cube. The volume of a cube with side length $s$ is given by:

$V_{cube} = s^3$

In this case, $s = 1.6'$, so:

$V_{cube} = (1.6)^3 = 4.096 \text{ ft}^3$

Next, we find the volume of the sphere. Since the sphere touches all 6 sides of the cube, the diameter of the sphere is equal to the side length of the cube. Therefore, the diameter of the sphere is 1.6', and the radius is half of that:

$r = \frac{1.6}{2} = 0.8 \text{ ft}$

The volume of a sphere with radius $r$ is given by:

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

In this case, $r = 0.8'$, so:

$V_{sphere} = \frac{4}{3}\pi (0.8)^3 = \frac{4}{3}\pi (0.512) \approx 2.14466 \text{ ft}^3$

Finally, we find the difference between the volume of the cube and the volume of the sphere:

$V_{difference} = V_{cube} - V_{sphere} = 4.096 - 2.14466 = 1.95134 \text{ ft}^3$

Rounding to the nearest tenth, we get $1.95134 \approx 2.0$.

3. Final Answer

2.0 ft^3