A company that manufactures storage bins for grains made a drawing of a silo. The silo has a conical base, as shown below:

Question

The figure shows a silo shaped as a closed cylinder with a conical end. The diameter of the silo is 4 ft, the length of the cylindrical part is 8.5 ft, and the entire length of the silo is 13 ft.

Which of the following could be used to calculate the total volume of grains that can be stored in the silo? (4 points)

π(8.5ft)2(2ft) + one over threeπ(2ft)2(13ft − 8.5ft)

π(2ft)2(8.5ft) + one over threeπ(2ft)2(13ft − 8.5ft)

π(8.5ft)2(2ft) + one over threeπ(13ft − 8.5ft)2(2ft)

π(2ft)2(8.5ft) + one over threeπ(13ft − 8.5ft)2(2ft)

Answer 1

To calculate the total volume of the silo, we need to consider both the cylindrical and conical parts. Let's analyze the dimensions given:

• Diameter of the silo: 4 ft → Radius \( r = \frac{4}{2} = 2 \) ft
• Length of the cylindrical part: 8.5 ft
• Total length of the silo: 13 ft → Length of the conical part: \( 13 - 8.5 = 4.5 \) ft

1. Volume of the cylindrical part: The formula for the volume of a cylinder is given by: \[ V_{\text{cylinder}} = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height (length) of the cylinder.

   For our silo: \[ V_{\text{cylinder}} = \pi (2 \text{ ft})^2 (8.5 \text{ ft}) = \pi (4) (8.5) = 34\pi \text{ ft}^3 \]

2. Volume of the conical part: The formula for the volume of a cone is given by: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius and \( h \) is the height (length) of the cone.

   For our silo: \[ V_{\text{cone}} = \frac{1}{3} \pi (2 \text{ ft})^2 (4.5 \text{ ft}) = \frac{1}{3} \pi (4) (4.5) = \frac{18}{3}\pi = 6\pi \text{ ft}^3 \]

3. Total volume of the silo: Now we can combine both volumes: \[ V_{\text{total}} = V_{\text{cylinder}} + V_{\text{cone}} = 34\pi + 6\pi = 40\pi \text{ ft}^3 \]

Now we need to find the formula presented in the options that matches our calculations:

• Option 1: \( \pi(8.5 \text{ ft})^2(2 \text{ ft}) + \frac{1}{3}\pi(2 \text{ ft})^2(13 \text{ ft} - 8.5 \text{ ft}) \)

• Option 2: \( \pi(2 \text{ ft})^2(8.5 \text{ ft}) + \frac{1}{3}\pi(2 \text{ ft})^2(13 \text{ ft} - 8.5 \text{ ft}) \)

• Option 3: \( \pi(8.5 \text{ ft})^2(2 \text{ ft}) + \frac{1}{3}\pi(13 \text{ ft} - 8.5 \text{ ft})^2(2 \text{ ft}) \)

• Option 4: \( \pi(2 \text{ ft})^2(8.5 \text{ ft}) + \frac{1}{3}\pi(13 \text{ ft} - 8.5 \text{ ft})^2(2 \text{ ft}) \)

From the calculations we did, the correct method to calculate the total volume of the silo is similar to Option 2. The parts involved are as follows:

• The cylindrical volume is given correctly as \( \pi(2 \text{ ft})^2(8.5 \text{ ft}) \).
• The conical volume is captured correctly by \( \frac{1}{3}\pi(2 \text{ ft})^2(4.5 \text{ ft}) \), which corresponds to \( (13 \text{ ft} - 8.5 \text{ ft}) \).

Thus, the correct choice to calculate the total volume of the silo is:

Option 2: \( \pi(2 \text{ ft})^2(8.5 \text{ ft}) + \frac{1}{3}\pi(2 \text{ ft})^2(13 \text{ ft} - 8.5 \text{ ft}) \).