Three Tennis Balls in a Cylindrical Can
Calculating Aluminum Required for Cylindrical Can
To create the can, approximately 879.6 cm² of aluminum is needed and the can will have about 118.0 cm³ of empty space.
Explanation:
To solve part (a) of your question, we first need to calculate the area of the base (which is a circle), and the volume of the cylinder that would consist of three stacked tennis balls. The radius of each ball (and the can) is half the diameter, which is 8.4 cm / 2 = 4.2 cm. Using the formula for the area of a circle (πr²), the area of the base of the can is approximately 55.42 cm². Since the height of the can is the diameter of three tennis balls, it is 8.4 cm x 3 = 25.2 cm. Therefore, the amount of aluminum needed (which is the surface area of the cylinder can) is calculated as 2πrh + 2πr² = 2(3.14)(4.2)(25.2) + 2(3.14)(4.2²) ≈ 879.6 cm². So, to the nearest tenth, you need about 879.6 cm² of aluminum to make the can.
Calculating Empty Space in the Can
To calculate the empty space in the can (part b), we compare the volume of the three balls (which is 3* 4/3πr³ ) and the volume of the cylinder can (which is πr²h). The difference is the amount of empty space in the can. Three tennis balls' total volume is 3 * 4/3 * 3.14 * (4.2³) ≈ 1236.0 cm³. The volume of the can is 3.14 * 4.2² * 25.2 ≈ 1118.0 cm³. Thus, the difference, to the nearest tenth, is about 118.0 cm³. So about 118.0 cm³ of the can is empty.
Three tennis balls that are each 8.4 cm in diameter are stacked in a cylindrical can.a) What is the minimum amount of aluminum required to make the cylindrical can?
b) If the three balls fit perfectly in the can, how much empty space is in the can?
(Round to the nearest tenth)
a) The minimum amount of aluminum required to make the cylindrical can is approximately 879.6 cm².
b) If the three balls fit perfectly in the can, there will be about 118.0 cm³ of empty space in the can.