You are building a sand castle and want to use a bucket that holds a volume of 826 in.cubed and has height 10.6 in. What is the radius of the​ bucket? Use 3.14 for pi. Use pencil and paper. If the height of the bucket is​ changed, but the volume stays the​ same, then how will the radius​ change? Explain.

1 answer

To find the radius of the bucket, we can use the formula for the volume of a cylinder, which is given by:

\[ V = \pi r^2 h \]

where:

  • \( V \) is the volume,
  • \( r \) is the radius,
  • \( h \) is the height, and
  • \( \pi \) is a constant (approximately 3.14).

We are given:

  • \( V = 826 , \text{in}^3 \)
  • \( h = 10.6 , \text{in} \)

We can substitute the values into the formula to solve for \( r \):

\[ 826 = 3.14 \cdot r^2 \cdot 10.6 \]

First, calculate \( 3.14 \cdot 10.6 \):

\[ 3.14 \cdot 10.6 = 33.324 \]

Now the equation becomes:

\[ 826 = 33.324 \cdot r^2 \]

Next, divide both sides by 33.324 to isolate \( r^2 \):

\[ r^2 = \frac{826}{33.324} \approx 24.76 \]

Now, take the square root to find \( r \):

\[ r \approx \sqrt{24.76} \approx 4.976 , \text{in} \]

So, the radius of the bucket is approximately \( 4.98 \text{ in} \).

How Radius Changes with Height

If the height of the bucket is changed but the volume remains the same, we can analyze how the radius will change.

From the volume formula:

\[ V = \pi r^2 h \]

If the volume \( V \) is kept constant and the height \( h \) changes, we can express the radius in terms of height:

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

From this equation, you can see that the radius \( r \) is inversely proportional to the square root of the height \( h \). This means:

  • If the height \( h \) increases, the radius \( r \) must decrease to maintain the same volume.
  • Conversely, if the height \( h \) decreases, the radius \( r \) must increase.

In summary, changing the height of the bucket while keeping the volume constant will lead to an inverse relationship between height and radius: as one increases, the other decreases.