Answer :
To determine the amount of plastic wrap needed to completely cover 8 cylindrical containers, we begin by calculating the area required for one container and then multiply by 8.
Step 1. Find the lateral surface area of one container.
The lateral surface area of a cylinder is given by:
[tex]$$
A_{\text{lateral}} = \pi \times \text{diameter} \times \text{height}.
$$[/tex]
Given that the diameter is [tex]$5.5$[/tex] inches and the height is [tex]$3$[/tex] inches, and using [tex]$\pi = 3.14$[/tex], we have:
[tex]$$
A_{\text{lateral}} = 3.14 \times 5.5 \times 3 \approx 51.8 \text{ in}^2.
$$[/tex]
Step 2. Find the area of the top and bottom (the two circular bases) of one container.
The area of a circle is:
[tex]$$
A_{\text{circle}} = \pi \times r^2,
$$[/tex]
where the radius is half the diameter. In this case:
[tex]$$
r = \frac{5.5}{2} = 2.75 \text{ inches}.
$$[/tex]
Thus, the area of one base is:
[tex]$$
A_{\text{circle}} = 3.14 \times (2.75)^2.
$$[/tex]
Since there are two bases (top and bottom), the combined area is:
[tex]$$
A_{\text{bases}} = 2 \times 3.14 \times (2.75)^2 \approx 47.5 \text{ in}^2.
$$[/tex]
Step 3. Calculate the total area for one container.
The total area required to completely wrap one container is the sum of the lateral surface area and the area of the two bases:
[tex]$$
A_{\text{container}} = A_{\text{lateral}} + A_{\text{bases}} \approx 51.8 \text{ in}^2 + 47.5 \text{ in}^2 = 99.3 \text{ in}^2.
$$[/tex]
Step 4. Compute the area required for 8 containers.
Multiply the area for one container by 8:
[tex]$$
A_{\text{total}} = 8 \times 99.3 \text{ in}^2 \approx 794.4 \text{ in}^2.
$$[/tex]
Thus, the minimum amount of plastic wrap needed to completely wrap 8 containers is approximately
[tex]$$\boxed{794.4 \text{ in}^2}.$$[/tex]
Step 1. Find the lateral surface area of one container.
The lateral surface area of a cylinder is given by:
[tex]$$
A_{\text{lateral}} = \pi \times \text{diameter} \times \text{height}.
$$[/tex]
Given that the diameter is [tex]$5.5$[/tex] inches and the height is [tex]$3$[/tex] inches, and using [tex]$\pi = 3.14$[/tex], we have:
[tex]$$
A_{\text{lateral}} = 3.14 \times 5.5 \times 3 \approx 51.8 \text{ in}^2.
$$[/tex]
Step 2. Find the area of the top and bottom (the two circular bases) of one container.
The area of a circle is:
[tex]$$
A_{\text{circle}} = \pi \times r^2,
$$[/tex]
where the radius is half the diameter. In this case:
[tex]$$
r = \frac{5.5}{2} = 2.75 \text{ inches}.
$$[/tex]
Thus, the area of one base is:
[tex]$$
A_{\text{circle}} = 3.14 \times (2.75)^2.
$$[/tex]
Since there are two bases (top and bottom), the combined area is:
[tex]$$
A_{\text{bases}} = 2 \times 3.14 \times (2.75)^2 \approx 47.5 \text{ in}^2.
$$[/tex]
Step 3. Calculate the total area for one container.
The total area required to completely wrap one container is the sum of the lateral surface area and the area of the two bases:
[tex]$$
A_{\text{container}} = A_{\text{lateral}} + A_{\text{bases}} \approx 51.8 \text{ in}^2 + 47.5 \text{ in}^2 = 99.3 \text{ in}^2.
$$[/tex]
Step 4. Compute the area required for 8 containers.
Multiply the area for one container by 8:
[tex]$$
A_{\text{total}} = 8 \times 99.3 \text{ in}^2 \approx 794.4 \text{ in}^2.
$$[/tex]
Thus, the minimum amount of plastic wrap needed to completely wrap 8 containers is approximately
[tex]$$\boxed{794.4 \text{ in}^2}.$$[/tex]
