High School

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 [tex]\pi = 3.14[/tex].

A. 51.8 in²
B. 99.3 in²
C. 595.8 in²
D. 794.4 in²

Answer :

The minimum amount of plastic wrap needed to completely wrap 8 cylindrical containers, each with a diameter of 5.5 inches and a height of 3 inches, is approximately 794.4 square inches. Therefore, the correct option is D.

To determine the minimum amount of plastic wrap needed to completely wrap 8 cylindrical containers, we must calculate the surface area of one container and then multiply it by 8. Given the diameter is 5.5 inches, the radius (r) is half of that, or 2.75 inches. The height (h) of the container is 3 inches.

The surface area (A) of a cylinder can be calculated using the formula

A = 2pi*r*h + 2pi*r^2,

where the first term represents the area of the side of the cylinder, and the second term is the area of the top and bottom circles (lids of the cylinder).

Plugging in our given values:

A = 2 * 3.14 * 2.75 * 3 + 2 *s 3.14 * (2.75)^2.

This gives us

A = 51.7 + 47.6

= 99.3 inches(^2) for one container. For 8 containers:

99.3 * 8 = 794.4 inches(^2).

Therefore, the minimum amount of plastic wrap needed to wrap 8 containers is approximately 794.4 square inches.