Question

A deli wraps its cylindrical containers of hot food items with plastic wrap. The containers have a diameter of 5.5 inches and a height of 3 inches. What is the minimum amount of plastic wrap needed to completely wrap 8 containers? Round your answer to the nearest tenth and approximate using π = 3.14.

Answer 1

Answer: 604.96 in²

Explanation:

Diameter (D) = 5.5 inches

Radius (r) = Diameter / 2 = 5.5 / 2 = 2.75 inches

Height (h) = 3 inches

Number of containers (n) = 8

To calculate the total surface area of one container:

Lateral Surface Area = 2 * π * r * h

Lateral Surface Area = 2 * 3.14 * 2.75 * 3 ≈ 51.93 square inches

Base Surface Area = π * r^2

Base Surface Area = 3.14 * (2.75)^2 ≈ 23.69 square inches

Total Surface Area of one container = Lateral Surface Area + Base Surface Area

Total Surface Area of one container ≈ 51.93 + 23.69 ≈ 75.62 square inches

Now, to find the total amount of plastic wrap needed to wrap all 8 containers:

Total amount of plastic wrap needed = 8 * Total Surface Area of one container

Total amount of plastic wrap needed ≈ 8 * 75.62 ≈ 604.96 square inches

Rounded to the nearest tenth, the minimum amount of plastic wrap needed to completely wrap 8 containers of hot food items is approximately 604.96 square inches. Hope that helps.