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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 \(\pi = 3.14\).

A. 51.8 in\(^2\)
B. 99.3 in\(^2\)
C. 595.8 in\(^2\)
D. 794.4 in\(^2\)

Answer :

To wrap 8 container 794.42 in² of plastic wrap required.

we have,

Diameter = 5.5 inch

radius= 5.5/2 inch

and, height = 3 inch

Surface area of Cylinder

= 2πrh + 2πr²

= 2 x 3.14 x 5.5/2 x 3 + 2 x 3.14 x 5.5/2 x 5.5/2

= 51.81 + 47.4925

= 99.3325 in²

Now, to wrap 8 container the amount of plastic wrap required

= 8 x 99.3325

= 794.42 in²

Learn more about Surface area here:

https://brainly.com/question/29298005

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