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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 794.4in². Therefore, the correct option is D.
To calculate the minimum amount of plastic wrap needed to completely wrap 8 cylindrical containers, we must first determine the surface area of one container and then multiply by 8, as each container would require wrapping individually. To do this, we'll use the formula for the lateral (side) surface area of a cylinder, which is A = 2 * π * r * h, where 'r' is the radius of the container and 'h' is the height.
The diameter of the container is given as 5.5 inches, so the radius (r) would be half of that, or 2.75 inches. The height (h) of the container is 3 inches. Plugging these values into the lateral surface area formula, we get:
A = 2 * π * 2.75in * 3in
A = 2 * 3.14 * 2.75in * 3in
A = 51.8in² for one container
To wrap 8 containers, we would multiply the area of one container by 8:
Total Plastic Wrap Required = 51.8in² * 8
Total Plastic Wrap Required = 414.4in²
However, this does not take into account the circular top and bottom of the cylinder, which also need to be covered with wrap. The area of a circle is calculated with the formula A = π * r². There are two circular ends on a cylinder, so we would double the result.
A (circular ends) = 2 * π * r²
A (circular ends) = 2 * 3.14 * (2.75in)²
A (circular ends) = 2 * 3.14 * 7.5625in²
A (circular ends) = 47.5in² for one container
For 8 containers, we need:
Total Circular Ends Plastic Wrap = 47.5in² * 8
Total Circular Ends Plastic Wrap = 380in²
To find the total amount of plastic wrap needed for 8 containers, we must add the wrap required for the lateral surfaces and the circular ends:
Total Plastic Wrap Required = Lateral Plastic Wrap + Circular Ends Plastic Wrap
Total Plastic Wrap Required = 414.4in² + 380in²
Total Plastic Wrap Required = 794.4in²
Therefore, the minimum amount of plastic wrap needed to completely wrap 8 containers is 794.4in², which should be rounded to the nearest tenth as per instruction, but in this case, we already have a one decimal place precision.
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