Example 3
Completion requirements
| Example 3 |
Multimedia |
(Video in Development)
The volume of canola piled on the ground in the shape of a cone is \(2\thinspace 000 \thinspace \rm {ft}^3\). The height of the pile is \(30 \thinspace \rm{ft}\). How wide is the base of the pile, given the formula for the volume of a cone is \(V = \frac{1}{3}\pi r^2h\)? Round to the nearest foot.
Step 1: Write down what you are given.
\(\begin{align}
V &= 2\thinspace 000 \thinspace {\rm{ ft}}^3 \\
h &= 30 \thinspace {\rm{ ft}} \\
r &= ? \\
\end{align}\)
Step 2: Substitute given values into the formula, and solve for \(r\).
\(\begin{align}
V &= \frac{1}{3}\pi r^2 h \\
2\thinspace 000 &= \frac{1}{3}\pi r^2 \left( {30} \right) \\
2\thinspace 000 &= 10\pi r^2 \\
\frac{{200}}{\pi } &= r^2 \\
\pm \sqrt {\frac{{200}}{\pi }} &= r \\
\pm 7.978... &= r \\
\end{align}\)
Because width cannot be negative, you can ignore the negative solution value.
Step 3: Determine the width of the pile.
To find the width, multiply the radius by two.
width = \(2(7.978...) = 15.957...\)
The width of the base of the pile is approximately \(16 \rm \thinspace {ft}\).
\(\begin{align}
V &= 2\thinspace 000 \thinspace {\rm{ ft}}^3 \\
h &= 30 \thinspace {\rm{ ft}} \\
r &= ? \\
\end{align}\)
Step 2: Substitute given values into the formula, and solve for \(r\).
\(\begin{align}
V &= \frac{1}{3}\pi r^2 h \\
2\thinspace 000 &= \frac{1}{3}\pi r^2 \left( {30} \right) \\
2\thinspace 000 &= 10\pi r^2 \\
\frac{{200}}{\pi } &= r^2 \\
\pm \sqrt {\frac{{200}}{\pi }} &= r \\
\pm 7.978... &= r \\
\end{align}\)
Because width cannot be negative, you can ignore the negative solution value.
Step 3: Determine the width of the pile.
To find the width, multiply the radius by two.
width = \(2(7.978...) = 15.957...\)
The width of the base of the pile is approximately \(16 \rm \thinspace {ft}\).