How many hockey balls can fit into CONTAINER A and CONTAINER B?

The diameter of CONTAINER A is $73 \text{ mm}$. The diameter of one hockey ball is $71 \text{ mm}$.
Since $73 \text{ mm} > 71 \text{ mm}$, only one hockey ball can fit across the diameter of the container.

The height of CONTAINER A is $428 \text{ mm}$.
Number of balls vertically = $\frac{\text{Height of CONTAINER A}}{\text{Diameter of one hockey ball}}$

Since only whole balls can fit, we take the floor of this value.

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 $L = 240 \text{ mm}$
Width $W = 17 \text{ cm} = 170 \text{ mm}$ (converting to mm)
Height $H = 100 \text{ mm}$

Number of balls along length = $\lfloor \frac{L}{d_{ball}} \rfloor = \lfloor \frac{240 \text{ mm}}{71 \text{ mm}} \rfloor = \lfloor 3.38 \rfloor = 3$
Number of balls along width = $\lfloor \frac{W}{d_{ball}} \rfloor = \lfloor \frac{170 \text{ mm}}{71 \text{ mm}} \rfloor = \lfloor 2.39 \rfloor = 2$
Number of balls along height = $\lfloor \frac{H}{d_{ball}} \rfloor = \lfloor \frac{100 \text{ mm}}{71 \text{ mm}} \rfloor = \lfloor 1.40 \rfloor = 1$

Total number of balls = (balls along length) $\times$ (balls along width) $\times$ (balls along height)

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 $\text{mm}^2$. Show, by means of calculations, if the worker is correct or incorrect.

Note: The formula provided for the surface area of a cylinder ($A = \pi \times r^2 \times h$) 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 $A = 2\pi r^2 + 2\pi r h$.

Calculate the surface area of CONTAINER A (Cylinder):
Radius $r_A = \frac{73 \text{ mm}}{2} = 36.5 \text{ mm}$
Height $h_A = 428 \text{ mm}$
$\pi = 3.14$

Calculate the surface area of CONTAINER B (Rectangular Prism):
Length $L_B = 240 \text{ mm}$
Width $B_B = 170 \text{ mm}$
Height $H_B = 100 \text{ mm}$

Surface area of CONTAINER A = $106607.11 \text{ mm}^2$
Surface area of CONTAINER B = $163600 \text{ mm}^2$

Since $106607.11 \text{ mm}^2 < 163600 \text{ mm}^2$, 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.