Question

Roberto is making sphere-shaped containers to ship inside a box. He has calculated that the volume of each sphere-shaped container can be no more than 530 cubic units. What is the largest whole-number dimension he can have as the radius of each sphere? Use 3.14 for pi.(1 point)Responses25 units25 units10 units10 units3 units3 units5 units

Answer 1

To determine the largest whole-number radius that Roberto can have for each sphere-shaped container, we start with the formula for the volume of a sphere:

\[
V = \frac{4}{3} \pi r^3
\]

Where \( V \) is the volume and \( r \) is the radius.

Given that the volume can be no more than 530 cubic units, we set up the inequality:

\[
\frac{4}{3} \pi r^3 \leq 530
\]

Substituting \( \pi \) with 3.14, we have:

\[
\frac{4}{3} (3.14) r^3 \leq 530
\]

Calculating \( \frac{4}{3} \times 3.14 \):

\[
\frac{4 \times 3.14}{3} = \frac{12.56}{3} \approx 4.1867
\]

So the inequality becomes:

\[
4.1867 r^3 \leq 530
\]

Now, divide both sides by 4.1867:

\[
r^3 \leq \frac{530}{4.1867} \approx 126.5
\]

Next, we take the cube root to find the maximum radius \( r \):

\[
r \leq 126.5^{1/3}
\]

Calculating \( 126.5^{1/3} \):

\[
\sqrt[3]{126.5} \approx 5
\]

Since we are looking for the largest whole number, we check \( r = 5 \):

Calculating the volume for \( r = 5 \):

\[
V = \frac{4}{3} \pi (5)^3 = \frac{4}{3} \cdot 3.14 \cdot 125 \approx \frac{4 \cdot 392.5}{3} \approx \frac{1570}{3} \approx 523.33
\]

This volume (approximately 523.33) is less than 530. Now let's check \( r = 6 \):

Calculating the volume for \( r = 6 \):

\[
V = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \cdot 3.14 \cdot 216 \approx \frac{4 \cdot 678.24}{3} \approx \frac{2712.96}{3} \approx 904.32
\]

This volume (approximately 904.32) exceeds 530.

Thus, the maximum whole-number dimension for the radius \( r \) is:

\[
\boxed{5 \text{ units}}
\]