To find the dimensions of the box that minimize the amount of material used, we can go through the problem step-by-step.

1. Step 1: Write the Volume Formula

   Given: $V = x^2 \cdot h = 296352 \, \text{cm}^3$ Here, $x$ is the length of one side of the square base, and $h$ is the height of the box.

   Rearrange to express $h$ in terms of $x$: $h = \frac{296352}{x^2}$

2. Step 2: Write the Surface Area Formula

   Since the box has an open top, the surface area $A$ consists of the base and the four sides: $A(x) = x^2 + 4 \cdot (x \cdot h)$ Substitute $h = \frac{296352}{x^2}$ into the formula: $A(x) = x^2 + 4 \cdot \left( x \cdot \frac{296352}{x^2} \right)$ Simplify this to: $A(x) = x^2 + \frac{1185408}{x}$

3. Step 3: Find the First Derivative $A'(x)$

   Differentiate $A(x)$ with respect to $x$: $A'(x) = 2x - \frac{1185408}{x^2}$

4. Step 4: Set the First Derivative to Zero to Find Critical Points

   Set $A'(x) = 0$: $2x - \frac{1185408}{x^2} = 0$ Multiply both sides by $x^2$ to clear the fraction: $2x^3 = 1185408$ Solve for $x$: $x^3 = \frac{1185408}{2} = 592704$ $x = \sqrt[3]{592704} \approx 84$

5. Step 5: Use the Second Derivative Test

   Find the second derivative $A''(x)$: $A''(x) = 2 + \frac{2370816}{x^3}$ Evaluate $A''(x)$ at $x = 84$: $A''(84) = 2 + \frac{2370816}{84^3}$ Since $A''(84) > 0$, the surface area is minimized at $x = 84$.

6. Step 6: Calculate the Height

   Substitute $x = 84$ back into the height formula: $h = \frac{296352}{84^2} = \frac{296352}{7056} \approx 42$

Therefore, the dimensions that minimize the material used are:

• Base dimensions: $84 \, \text{cm} \times 84 \, \text{cm}$
• Height: $42 \, \text{cm}$

Would you like more details on any step?

Here are five questions that expand on this solution:

1. How does changing the volume of the box affect the optimal dimensions for minimizing surface area?
2. What if the box had a closed top—how would that change the surface area formula?
3. Can we generalize this method to find the minimum material for boxes of different shapes?
4. Why does the second derivative test confirm that $A(x)$ has a local minimum at $x = 84$?
5. How would the solution change if the volume were given in different units?

Tip: When dealing with optimization problems, always check if you need to maximize or minimize the function and remember to use derivatives to find critical points and confirm their nature.