Let's go through the solution step-by-step.

Given Information

1. The surface area of a cubic cake is $16 \, \text{cm}^2$.
2. We need to find the maximum number of these cakes that can fit inside a cubic container with side length $15 \, \text{cm}$.

Step 1: Find the Side Length of Each Cake

Since the surface area $A$ of a cube is given by: $A = 6s^2$ where $s$ is the side length of the cube.

We can solve for $s$: $6s^2 = 16$ $s^2 = \frac{16}{6} = \frac{8}{3} \approx 2.67$ $s \approx \sqrt{2.67} \approx 1.63 \, \text{cm}$

So, each cake has a side length of approximately $1.63 \, \text{cm}$.

Step 2: Find the Volume of the Container and Each Cake

1. Volume of the container (with side length $15 \, \text{cm}$): $V_{\text{container}} = 15^3 = 3375 \, \text{cm}^3$

2. Volume of one cake (with side length $1.63 \, \text{cm}$): $V_{\text{cake}} = (1.63)^3 \approx 4.33 \, \text{cm}^3$

Step 3: Calculate the Maximum Number of Cakes

To find the maximum number of cakes that can fit inside the container, divide the volume of the container by the volume of one cake: $\text{Number of cakes} = \frac{3375}{4.33} \approx 779.91$

Since we can't have a fraction of a cake, we take the integer part: $\text{Maximum number of cakes} = 779$

Final Answer

The maximum number of cakes that can fit in the container is 779.

Would you like further clarification on any step, or a detailed breakdown?

Here are some related questions to deepen your understanding:

1. How would the solution change if the container was not cubic but rectangular?
2. What if the surface area of each cake were different, such as 18 cm²? How would that impact the result?
3. How could we calculate the packing efficiency if cakes were not perfect cubes?
4. How would changing the units (e.g., from cm to m) impact our solution approach?
5. If the cakes were spheres instead of cubes, how would we determine the maximum number?

Tip: Remember, for volume-based problems, understanding the basic shape formulas helps simplify the calculation process significantly.