A can of Pringles has a height of 23.2 cm, a radius of 3.8 cm, a circumference of 23.9 cm, and a volume of 1052.5 cm. a. determine the maximum optimized volume of the can b. calculate how volume optimized the can is, i.e. the % error c. make a graph of the can’s volume versus the can’s radius using Desmos (plot one point for the maximum possible volume and plot one point for the can’s actual volume of 1052.5 cm).
A can of Pringles has a height of 23.2 cm, a radius of 3.8 cm, a circumference of 23.9 cm, and a volume of 1052.5 cm.
a. determine the maximum optimized volume of the can
b. calculate how volume optimized the can is, i.e. the % error
c. make a graph of the can’s volume versus the can’s radius using Desmos (plot one point for the maximum possible volume and plot one point for the can’s actual volume of 1052.5 cm).
The Correct Answer and Explanation is :
a. Determine the maximum optimized volume of the can
The maximum volume of a cylindrical can is obtained when its dimensions are optimized. For a cylinder, the volume formula is:
[
V = \pi r^2 h
]
Where:
- ( V ) is the volume,
- ( r ) is the radius, and
- ( h ) is the height.
To optimize the volume of the can, we are assuming that the given radius and height are the maximum possible values that are allowed under the constraints. The height of the can is given as 23.2 cm, and the radius can be calculated using the formula for circumference:
[
C = 2\pi r
]
We are given the circumference ( C = 23.9 \, \text{cm} ). Solving for the radius:
[
r = \frac{C}{2\pi} = \frac{23.9}{2\pi} \approx 3.8 \, \text{cm}
]
Now, using the radius ( r = 3.8 \, \text{cm} ) and height ( h = 23.2 \, \text{cm} ), we can calculate the maximum volume:
[
V_{\text{max}} = \pi (3.8)^2 (23.2) \approx 3.1416 \times 14.44 \times 23.2 \approx 1052.5 \, \text{cm}^3
]
This result indicates that the maximum optimized volume of the can is 1052.5 cm³, which matches the given volume. Therefore, the can is already optimized.
b. Calculate how volume optimized the can is (i.e., the % error)
The volume optimized can is the same as the actual volume, meaning the calculated volume and the actual volume are identical. To calculate the percentage error, we use the formula:
[
\text{Percentage Error} = \left( \frac{\left| \text{Theoretical Volume} – \text{Actual Volume} \right|}{\text{Theoretical Volume}} \right) \times 100
]
Substituting the values:
[
\text{Percentage Error} = \left( \frac{\left| 1052.5 – 1052.5 \right|}{1052.5} \right) \times 100 = 0\%
]
Thus, there is no error in the volume, and the can is 100% volume-optimized.
c. Graph the can’s volume versus the can’s radius using Desmos
You can plot the volume of a cylinder as a function of its radius by using the formula ( V = \pi r^2 h ). Here, the height is constant at 23.2 cm. In Desmos, you would input:
[
V(r) = \pi r^2 \times 23.2
]
Then, plot one point for the optimized volume with ( r = 3.8 \, \text{cm} ) and another point for the actual volume of 1052.5 cm³.
You would plot the following points:
- Optimized volume: ( (3.8, 1052.5) )
- Actual volume: ( (3.8, 1052.5) )
Since both the optimized and actual volume are the same, both points will coincide on the graph. The relationship between radius and volume will show the parabolic increase in volume as the radius increases.