This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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It looks like we're moving on to question 2.3. Let's tackle these packaging problems! First, let's determine the dimensions of a hockey ball: Radius of a hockey ball r_ball = 35.5 mm. Diameter of a hockey ball d_ball = 2 × r_ball = 2 × 35.5 mm = 71 mm. 2.3.1 How many hockey balls can fit into CONTAINER A? CONTAINER A is a cylinder with a diameter of 73 mm and a height of 428 mm. Step 1: Determine how many balls can fit across the diameter. The diameter of CONTAINER A is 73 mm. The diameter of one hockey ball is 71 mm. Since 73 mm > 71 mm, only one hockey ball can fit across the diameter of the container. Step 2: Determine how many balls can fit vertically along the height. The height of CONTAINER A is 428 mm. Number of balls vertically = Height of CONTAINER ADiameter of one hockey ball Number of balls = 428 mm71 mm ≈ 6.028 Since only whole balls can fit, we take the floor of this value. Number of balls = 6.028 = 6 Therefore, 6 hockey balls can fit into CONTAINER A. 2.3.2 How many hockey balls can fit into CONTAINER B? CONTAINER B is a rectangular prism with dimensions: Length L = 240 mm Width W = 17 cm = 170 mm (converting to mm) Height H = 100 mm Step 1: Calculate how many balls fit along each dimension. Number of balls along length = (L)/(d_ball) = 240 mm71 mm = 3.38 = 3 Number of balls along width = (W)/(d_ball) = 170 mm71 mm = 2.39 = 2 Number of balls along height = (H)/(d_ball) = 100 mm71 mm = 1.40 = 1 Step 2: Calculate the total number of balls. Total number of balls = (balls along length) × (balls along width) × (balls along height) Total balls = 3 × 2 × 1 = 6 Therefore, 6 hockey balls can fit into CONTAINER B. 2.3.3 Explain why CONTAINER A is better for packaging the hockey balls. Both containers can fit the same number of hockey balls (6 balls). However, CONTAINER A is a cylinder, which is a better shape for packaging spherical objects like hockey balls. A cylindrical container minimizes the empty space (voids) around the curved surfaces of the balls compared to a rectangular prism, leading to more efficient use of packaging material and potentially better protection against movement. 2.3.4 A worker tells the supplier that CONTAINER B has the smaller surface area in mm^2. Show, by means of calculations, if the worker is correct or incorrect. Note: The formula provided for the surface area of a cylinder (A = × r^2 × h) is actually the formula for the volume of a cylinder. I will use the correct formula for the surface area of a cylinder, which is A = 2 r^2 + 2 r h. Calculate the surface area of CONTAINER A (Cylinder): Radius r_A = 73 mm2 = 36.5 mm Height h_A = 428 mm = 3.14 Step 1: Use the correct surface area formula for a cylinder. A_A = 2 r_A^2 + 2 r_A h_A Step 2: Substitute the values and calculate. A_A = 2 × 3.14 × (36.5 mm)^2 + 2 × 3.14 × 36.5 mm × 428 mm A_A = 2 × 3.14 × 1332.25 mm^2 + 2 × 3.14 × 15638 mm^2 A_A = 8360.47 mm^2 + 98246.64 mm^2 A_A = 106607.11 mm^2 Calculate the surface area of CONTAINER B (Rectangular Prism): Length L_B = 240 mm Width B_B = 170 mm Height H_B = 100 mm Step 1: Use the given surface area formula for a rectangular prism. A_B = 2(L_B × B_B) + 2(L_B × H_B) + 2(B_B × H_B) Step 2: Substitute the values and calculate. A_B = 2(240 mm × 170 mm) + 2(240 mm × 100 mm) + 2(170 mm × 100 mm) A_B = 2(40800 mm^2) + 2(24000 mm^2) + 2(17000 mm^2) A_B = 81600 mm^2 + 48000 mm^2 + 34000 mm^2 A_B = 163600 mm^2 Step 3: Compare the surface areas. Surface area of CONTAINER A = 106607.11 mm^2 Surface area of CONTAINER B = 163600 mm^2 Since 106607.11 mm^2 < 163600 mm^2, CONTAINER A has a smaller surface area than CONTAINER B. Therefore, the worker's statement that CONTAINER B has the smaller surface area is incorrect. That's 2 down. 3 left today — send the next one.